(−2 1/5) × (−5 3/4) as a fraction

Answer:

That would be negative 45 or - 45

Please correct me if im wrong

Answer:

Hundreds : 1

Tens : 6

Ones : 2

.

Tenths : 8

Hundredths : 9

Thousandths : 1

Answer:

X equals positive -1.75

Step-by-step explanation:

because X is muliplying by the common number attached to it (4)

-1.75 times 4 equals -7

thus being -7-85= -92

Answer:

well you have to solve for X  so -1.75 times 4 equals -7 then solve. so your answer would be- X equals positive -1.75

Answer: $113.89

Step-by-step explanation:

108.99 * 0.045

= 4.90455

108.99 + 4.90455

= 113.89455

$113.89

Answer:

Total Cost = $113.89

Step-by-step explanation:

Total Cost = Cell phone cost + sale Tax % ( cell phone cost)

Total Cost = $108. 99 + .045 × $108.99

Total Cost = $108.99 + $4.90455

Total Cost = $108.99 + $4.90

Total Cost = $113.89

The correct word form of 8.16 is b. eight and sixteen hundredths

Since the number is 8.16, we have 8.16 = 8 + 0.16 = 8 + 16/100

Since we have 8 in the units place, it is pronounced eight.

Also, we have our decimal part 0.16 = 16/100 which is pronounced as sixteen hundredths.

So, combining both we have eight and sixteen hundredths.

So, the correct word form of 8.16 is b. eight and sixteen hundredths

Learn more about decimals here:

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Answer:

(2,11)

Step-by-step explanation:

what i did was actually substitution

It's like ..what I do when I've forgotten how to solve a particular equation in math or even physics

But I'd recommend you find the real solution cuz there's a formula

Answer:

7 boys to 10 all children

Answer:

16

Step-by-step explanation:

y = 7

4(7)-12

28-12 = 16

Answer:

[tex]-\frac{9}{r^4}[/tex]

Step-by-step explanation:

Using Power rule

[tex]\frac{d}{dr} (r^n)=nr^{n-1}\\-\frac{9}{r^4}[/tex]

The answer is F'(r) = [tex]\mathrm {-\frac{9}{r^{4}}}[/tex].

We can rewrite the function as :

F(r) = 3r⁻³

Now, differentiate by power rule :

F'(r) = (-3)3r⁻³⁻¹

F'(r) = -9r⁻⁴

F'(r) = -9/r⁴

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation that goes through (5,4) and (10,2)

[tex](\stackrel{x_1}{5}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{2}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{2}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{10}-\underset{x_1}{5}}}\implies \cfrac{-2}{5} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-\cfrac{2}{5}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{5}{2}}\qquad \stackrel{negative~reciprocal}{+\cfrac{5}{2}\implies \cfrac{5}{2}}}[/tex]

Answer:

Measure of Angle B = 25°

Step-by-step explanation:

Something to keep in mind: Angle A and B are Vertical Angles so they will always be equal

so;

Angle A = Angle B

(4x + 5) = (-2(x - 17.5))

4x + 5 = -2x + 35   (<= Solved by distributive property)

6x + 5 = 35

6x = 30

x = 5

Now we can find the measure of Angle B, by substituting x by 5 in the B equation;

-2 (x - 17.5) = Angle B

-2 (5 - 17.5) = Angle

-2 (-12.5) = Angel B

25 = Angle B

so measure of angle B = 25

Hope this helps!

The measures of the angles and the lengths of the sides of the triangle can

be found using sine rule.

The correct responses are;

Reasons:

The given parameter are;

∠B = 65°20', b = 5.38 inches, c = 4.88 inches

∠C ≥ 90°

By sine rule, we have;

[tex]\dfrac{5.38}{sin(65^{\circ} 20')} = \mathbf{ \dfrac{4.88}{sin(\angle C)}}[/tex]

[tex]sin(\angle C) = \mathbf{ \dfrac{sin(65^{\circ} 20') \times 4.88}{5.38}}[/tex]

[tex]\angle C = arcsin \left(\dfrac{sin(65^{\circ} 20') \times 4.88}{5.38} \right) \approx \mathbf{ 55.517 ^{\circ}}[/tex]

∠C = 55.517° < 90°, therefore, the correct response is; None

[tex]\dfrac{sin(65^{\circ} 20')}{5.38} = \dfrac{4.88}{sin(\angle C)}[/tex]

∠A = 180° - 65°20' - 55.517° = 59.15° = 59°9'

∠A = 59°9'

Therefore;

[tex]\dfrac{5.38}{sin(65^{\circ} 20')} = \mathbf{ \dfrac{a}{sin(59.15^{})}}[/tex]

[tex]a = \dfrac{5.38 \times sin(59.15^{\circ}) }{sin(65^{\circ} 20')} \approx \mathbf{ 5.083}[/tex]

a ≈ 5.083 inches

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Answer:

4.4 hours

Step-by-step explanation:

It would take Shawn 4.4 hours, as this would give a time of .4 hours to Michelle, totaling 4.8 hours

Answer:

Step-by-step explanation:

4) parallel because 118° is a supplement to 62° and the corresponding angles are both 118°

5) NOT parallel. The labeled angles sum to 120° and would sum to 180° for parallel lines.

6) NOT parallel. see pic.

If parallel, extending a line to intersect ℓ₁ makes an opposite internal angle which would also be 48°. The created triangle would have its third angle at 180 - 90 - 48 = 42° which is opposite a labeled 48° angle, which is false, so the lines cannot be parallel

7)

b = 78°  as it corresponds with a labeled angle above it

a = 180 - 78 = 102° as angles along a line from a common vertex sum to 180

f = is an opposite angle to 180 - 78 - 44 = 58° as angles along a line from a common vertex sum to 180

e = 180 - 90 - 64 = 26° as angles along a line from a common vertex sum to 180

c = 58° as it corresponds with f

d = 180 - 58 = 122° as angles along a line from a common vertex sum to 180

integers is your answer

Answer: $63.03

Step-by-step explanation:

$67.82 - $4.79 = 63.03

Answer:

8 years = 5600

Step-by-step explanation:

10 years = 3500x2= 7000

1 year = 7000/10 = 700

1 year = 700

8 years = 700x8 = 5600

Answer:

1hr 5 minutes

Step-by-step explanation:

9hrs 5mins-8 hrs=1 hr 5 mins

mark me brainliest plz if you are helped

Answer:

-5x + 3

Step-by-step explanation:

1/3 x - 4x -5/3x + 1/3 x +3

2/3 x -4 -5/3x + 3

3/3x -4x +3

-1 1x -4x + 3

- 5x + 3

:)

Answer:

Your tip is $15.00

Step-by-step explanation:

Have a nice day ^^

Answer:

w = 22 in, h = 8 in.

Step-by-step explanation:

Reading the problem, I immediately see two equations. This hints me towards a system of equations problem.

Statement 1 is The width of a triangle is six more than twice the height.

The width (w) of a triangle is (=) six (6) more (+) than twice (2) the height (h).

Equation 1: w = 6 + 2h

Statement 2 is The area of the triangle is 88 in2.

The formula for the area of a triangle is [tex]\frac{wh}{2}[/tex]. So, we get our second equation.

Equation 2: wh/2 = 88

We see that we have the same number of equations as unknowns, two. This is important as it lets us know that we can find two unknowns in the two equations. If there are more unknowns than equations, then we cannot find the values of all the variables. Look for another equation in the problem. However, if the # equations greater than or equal to number of unknowns, then we are ready to start solving!  So, we can now move to solving.

Equation 1 isolates one of the unknowns (w). So, we can replace w in the second equation with the right side of eq 1. Then, we simplify for h.

[tex]\frac{wh}{2} = 88[/tex]

[tex]\frac{(6 + 2h)h}{2} = 88[/tex]

[tex]\frac{6h + 2h^2}{2} = 88[/tex]

[tex]6h + 2h^2 = 176[/tex]

[tex]2h^2 + 6h - 176 = 0[/tex]

[tex]h = -11, 8[/tex] (assuming you know how to solve quadratics)

Since we know that heights of triangles cannot be negative, we find that [tex]h = 8[/tex] in.

For the final step, we plug [tex]h[/tex] into our first equation to find [tex]w[/tex].

[tex]w = 6 + 2h[/tex]

[tex]w = 6 + 2(8)[/tex]

[tex]w = 6 + 16[/tex]

[tex]w = 22[/tex]

So, the answer is w = 22 in, h = 8 in.

To check if we were right, we can plug the values into both equations to see if they are equal. I will skip equation 1 since we used that equation "as-is" to find w.

[tex]\frac{wh}{2} = 88[/tex]

[tex]\frac{22 * 8}{2} = 88[/tex]

[tex]11 * 8 = 88[/tex]

[tex]88 = 88[/tex]

The solution checks out!

Answer:

Step-by-step explanation:

Arc angle KLM will be twice ∠KSM = 210°

Arc angle ML will be 210 - 130 = 80°

Arc angle LMS will be twice ∠LKS = 160°

Arc angle SM will be Arc angle LMS - ML = 160 - 80 = 80°

Answer: IJ = 15

J

15. x

H. I

14

Two congruent Angles —-> Isosceles Triangle

Opposite sides must be congruent.

J

15 15

H. I

14

Answer:

The most common benchmark percents are 0%, 10%, 25%, 50%, 75% and 100%.

To find 60% of 280 we first find 10% of 280 which is 28. 6 x 28 = 168.

To find 60% of 280 we first find 50% of 280 which is 140. Now we add 10% which we calculated before: 140 + 28 = 168.

Answer:

Step-by-step explanation:

Explain two ways you can benchmark percents to find 60% of 280

1) 280 : 100 * 60 = 168

2) 280 * 0.6 = 168

Answer:

  3

Step-by-step explanation:

5(9/12 -1/4) +0.5

= 5(9/12 -3/12) +1/2 . . . . . . express the fractions using a common denominator

= 5(6/12) +1/2 . . . . . . . add the fractions

= 5(1/2) +1/2 . . . . . . reduce the result

= (5 +1)(1/2) . . . . factor out 1/2 (easier than adding mixed numbers)

= 6/2 = 3

_____

A calculator can be a useful tool for evaluating numerical expressions.

Answer:

36.11 feet

Step-by-step explanation:

If the circumfrence goes across the circle, and the radius goed halfway, substitute the measuremants. C/2. So, divide 72.22 by 2 and you get 36.11 feet!

It takes Brian 1.5 hours to get home from school

Speed is the ratio of distance travelled to time taken. It is given by:

Speed = distance / time

Let d represent the distance from their school to the house and t represent the time (hours) it takes Brian to teach home.

For Brian:

4 = d/t

d = 4t

For Scott:

6 = d / (t - 1/2)

d = 6t - 3

4t = 6t - 3

2t = 3

t = 1.5 hours

Therefore it takes Brian 1.5 hours to get home from school

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Step-by-step explanation:

[tex]\small{ \displaystyle\tt \int_{0}^{1} \int_{0} ^{ x_{1}} \int_{ 0 }^{x_{2}}... \int_{ 0 }^{x_{99}} ln(1 -x_{100} ) dx_{100}... dx_{1}}[/tex]

[tex]\small{{ \displaystyle\tt = - \int_{0}^{1} \int_{0} ^{ x_{1}} \int_{ 0 }^{x_{2}}... \int_{ 0 }^{x_{98}} \sum_{n = 1}^{ \infty \frac{}{} \frac{}{} } \frac{1}{n} \int_{0}^{x_{99}} {x}^{n}_{100} \: dx_{100}...dx_{1} }}[/tex]

[tex]{{ 0 }^{x_{98}} \frac{{x}^{n + 1}_{99}}{n + 1} \: dx_{99}...dx_{1} }[/tex]

[tex]\small{ \displaystyle\tt = -\sum_{n = 1}^{ \infty \frac{}{} \frac{}{} } \frac{1}{n}\int_{0}^{1} \int_{0} ^{ x_{1}} }[/tex]

[tex]\displaystyle \tt= - \frac{1}{(100)!_{} } \sum_{n = 1}^{ \infty } \frac{(n - 1)!(100)!}{(n + 101)!}=−(100)!1n=1∑∞(n+101)!(n−1)!(100)![/tex]

[tex]\displaystyle \tt= - \frac{1}{(100)!_{} } \sum_{n = 1}^{ \infty} \int_{0}^{1} {x}^{n - 1} (1 - x {)}^{100} [/tex]

[tex]\displaystyle \sf - \dfrac{1}{(100!)} \int_{0}^{1} \dfrac{(1 - x)^{100} }{1 - x} \:[/tex]

[tex]\displaystyle\tt { = - \frac{1}{(100)!} \int_{0}^{1} {x}^{99} \: dx=−(100)!1∫01x99dx}

[/tex]

Step-by-step explanation:

[tex]{ \pink{ \displaystyle\tt \int_{0}^{1} \int_{0} ^{ x_{1}} } \sf \pink\int_{ 0 }^{x_{2}... \int_{ 0 }^{x_{99}}} ln(1 -x_{100} ) dx_{100}... dx_{1}}[/t

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