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3 years ago

The length of a rectangle is 4 inches more than the width. the perimeter is 32 inches. find the length and width.

1 answer:

7 0

The length of the rectangle is 10 inches, and the width is 6 inches.

10 + 10 + 6 + 6 = 32

And that's your answer! ^ I hope this helped! c:

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When 0.3(4x-8)-0.5(-2.4x+4) is simplified what is the resulting expression

Fantom [35]

Answer:

2.4x-4.4

Step-by-step explanation:

The first step to simplify this expression is to expand the terms in parenthesis

3 0

3 years ago

Read 2 more answers

)The ratio of X's to O's is 15 over 10. . Use the image below to determine another ratio for X's to O's: Image with 15 Xs and 10

luda_lava [24]

Answer:

2:3

Step-by-step explanation:

The common factor between 10 and 15 is 5.

5 goes into 10 twice and 15 three times.

$\frac{15}{10}$ ÷ $\frac{5}{5} = \frac{2}{3}$

4 0

2 years ago

A trapezoid has a height of 8 cm and bases with lengths of 14 cm and 8 cm. Tina says the area of the trapezoid is 176 cm2.

Sonbull [250]

Answer:

Step-by-step explanation:

Let's do this properly:

Area of trapezoid = average length times height.

Here,

Area = (1/2)(8 cm + 14 cm)(8 cm), or

= (11 cm)(8 cm) = 88 cm^2

It appears that Tina forgot to divide the sum of 14 cm and 8 cm by 2. Thus, her answer is twice as large as it should be.

7 0

3 years ago

Read 2 more answers

Write the equation of the line in Point-Slope Form

katrin [286]

17.

The point-slope form:

$y-y_1=m(x-x_1)$

We have slope=m=8, and the point (-5, -3). Substitute:

$y-(-3)=8(x-(-5))\\\\\boxed{y+3=8(x+5)}$

18.

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (3, 4) and (15, 10). Substitute:

$m=\dfrac{10-4}{15-3}=\dfrac{6}{12}=\dfrac{1}{2}$

The point-slope form:

$\boxed{y-4=\dfrac{1}{2}(x-3)}$

$\boxed{y-10=\dfrac{1}{2}(x-15)}$

4 0

3 years ago

Show that (0, 0) is equidistant from (1, 2), (-1, -2) and (2, 1).

Len [333]

Step-by-step explanation:

1) if (0;0) if point A, (1;2) is point B; (-1;-2) is C, (2;1) is D, then

2) the vector AB is (1;2); vector AC is (-1;-2) and vector AD is (2;1), then

3) the length of AB, AC and AD are:

$|AB|=\sqrt{1^2+2^2}=\sqrt{5};$

$|AC|=\sqrt{(-1)^2+(-2)^2}=\sqrt{5};$

$|AD|=\sqrt{2^2+1^2}=\sqrt{5}.$

4) if the lengths above are equal, then it means that (0, 0) is equidistant from (1, 2), (-1, -2) and (2, 1).

3 0

2 years ago