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wariber [46]
1 year ago
11

Find the area round to two decimal places as needed

Mathematics
1 answer:
Mademuasel [1]1 year ago
3 0

To find the area of an obtuse triangle you have to multiply the base of the triangle by the vertical height and divide the result by 2 following the formula:

A=\frac{b\cdot h}{2}

The base of the triangle is b= 7 miles and the height is h= 8 miles, using these lengths calculate the area as follows:

\begin{gathered} A=\frac{7\cdot8}{2} \\ A=\frac{56}{2} \\ A=28mi^2 \end{gathered}

The area of the triangle is 28 square miles.

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A coupon subtracts $5.16 from the price (p) of a shirt. You pay $15.48 for the shirt after using the coupon. Write and solve an
lbvjy [14]

The equation would look like: p - 5.16 = 15.48

because the coupon subtracts 5.16 from p and the price you pay is 15.48

Solve the equation by adding 5.16 to both sides

p = 20.64

The original price of the shirt is $20.64

7 0
3 years ago
Sasha is cutting a cake into 42 equal-sized pieces. If 14 people want cake, then they can each have 3 slices. And if 21 people w
kupik [55]

Answer:

k = 42

the constant of proportionality in this inverse variation is 42.

Step-by-step explanation:

Let x represent the number of people that want cake

And y the number of slices each can have;

Since x is inversely proportional to y;

x = k/y .....1

Where k is the proportionality constant;

To solve for k, we will substitute any of the case scenario given into equation 1.

If 14 people want cake, then they can each have 3 slices;

x = 14 , y = 3

substituting;

14 = k/3

k = 14 × 3 = 42

k = 42

the constant of proportionality in this inverse variation is 42.

Therefore,

x = 42/y

4 0
3 years ago
3 ounces of cinnamon cost $2.40. There are 16 ounces in 1 pound how much the cinnamon cost per pound
faltersainse [42]
16 ounces in a pound
2.40 for 3 ounces
16 divided by 3 = 5.333
2.40 * 5.333 =
12.8
Approx $12.80 for 1 pound of cinnamon
3 0
3 years ago
Show that cos(3x) = cos3(x) − 3 sin2(x)cos(x). (Hint: Use cos(2x) = cos2(x) − sin2(x) and the cosine sum
otez555 [7]

<u>Step-by-step explanation:</u>

To prove:

\cos 3x=\cos^3x-3\sin^2 x\cos x

Identities used:

\cos(A+B)=\cos A\cos B+\sin A\sin B      ......(1)

\sin 2A=2\sin A\cos A        ........(2)

\cos 2A=\cos^2A-\sin^2 A     .......(3)

Taking the LHS:

\Rightarrow \cos 3x=\cos (x+2x)

Using identity 1:

\Rightarrow \cos (x+2x)=\cos x\cos 2x-\sin x\sin 2x

Using identities 2 and 3:

\Rightarrow \cos x(\cos ^2x-\sin^2 x)-\sin x(2\sin x\cos x)\\\\\Rightarrow \cos^3x-\sin^2x\cos x-2\sin^2 x\cos x\\\\\Rightarrow \cos^3x-3\sin^2x\cos x

As, LHS = RHS

Hence proved

4 0
3 years ago
If the parallel sides of a trapezoid are contained by the lines and , find the equation of the line that contains the midsegment
Fittoniya [83]

Answer:

Equation of midsegment line: y = (-1/4)x + 2.

Step-by-step explanation:

If the parallel sides of a trapezoid are contained by the lines:-

y = (-1/4)x +5 and y = (-1/4)x - 1

Midsegment of any trapezoid is the line segment

1. that is parallel to pair of parallel side of trapezoid and

2. that passes through the middle of the trapezoid and cuts the other two sides into equal-half.

It means the midsegment would have same slope as the parallel lines and y-intercept would be in the middle of intercepts of parallel lines.

So y = mx + b

where m = -1/4 and b = (5 - 1)/2 = 4/2 = 2.

Hence, the equation of midsegment would be y = (-1/4)x + 2.

6 0
3 years ago
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