The ratio of 6 nickels to 4 dollars is

Find the area of the parallelogram shown below.

ale4655 [162]

Answer:

<h2>120 square units </h2>

Step-by-step explanation:

Given, 

Base of the parallelogram = 10

Height of the parallelogram = 12

Therefore, 

Area of the parallelogram = base × height 

= 10 × 12

= 120

Hence,

Area of the parallelogram is 120 square units (Ans)

5 0

3 years ago

Trigonometry

Alchen [17]

The area of the triangle PQR is 17.6 square units.

Explanation:

Given that the sides of the triangle are PQ = 12 and PR = 3 and $m\angle P = 78^{\circ}$

We need to determine the area of the triangle PQR

Area of the triangle:

The area of the triangle can be determined using the formula,

$\text {Area}=\frac{1}{2} qr \sin P$

Substituting the values, we get,

$\text {Area}=\frac{1}{2}(12)(3) \sin 78$

Simplifying, we have,

$\text {Area}=\frac{1}{2}(36)(0.98)$

Multiplying the terms, we have,

$\text {Area}=\frac{35.28}{2}$

Dividing, we get,

$\text {Area}=17.64$

Rounding off to the nearest tenth, we have,

$Area=17.6$

Thus, the area of the triangle PQR is 17.6 square units.

3 0

3 years ago

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Which of the following shows the graph of y = 2e*?

faust18 [17]

You didn’t add any pictures so how could i help you with this?

7 0

2 years ago

I don't know what the answer is.

Ksenya-84 [330]

Answer:

C.) 87

Step-by-step explanation:

if you look at it from the other side, its counting down from 91 from the left.

6 0

3 years ago

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Solve for y 16y^2-25=0

Pepsi [2]

Answer:

$\large\boxed{x=-\dfrac{5}{4}\ \vee\ x=\dfrac{5}{4}}$

Step-by-step explanation:

$16y^2-25=0\\\\METHOD\ 1:\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\16=4^2\ \text{and}\ 25=5^2\ \text{therefore we have}\\\\4^2y^2-5^2=0\\\\(4y)^2-5^2=0\\\\(4y-5)(4y+5)+0\iff4y-5=0\ \vee\ 4y+5=0\\\\4y-5=0\qquad\text{add 5 to both sides}\\4y=5\qquad\text{divide both sides by 4}\\\boxed{y=\dfrac{5}{4}}\\\\4y+5=0\qquad\text{subtract 5 from both sides}\\4y=-5\qquad\text{divide both sides by 4}\\\boxed{x=-\dfrac{5}{4}}$

$METHOD\ 2:\\\\16y^2-25=0\qquad\text{add 25 to both sides}\\\\16y^2=25\qquad\text{divide both sides by 16}\\\\y^2=\dfrac{25}{16}\to y=\pm\sqrt{\dfrac{25}{26}}\\\\y=-\dfrac{\sqrt{25}}{\sqrt{16}}\ \vee\ x=\dfrac{\sqrt{25}}{\sqrt{16}}\\\\\boxed{y=-\dfrac{5}{4}\ \vee\ x=\dfrac{5}{4}}$

7 0

3 years ago

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