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3 years ago

Please help with this question.

Mathematics

1 answer:

Anit [1.1K]3 years ago

4 0

The equation is

Area of a triangle = (1/2) · (base) · (height)

Plug in the numbers given in the question:

52 square inches = (1/2) · (base) · (13 inches)

Divide each side by (13 inches):

4 inches = (1/2) · (base)

Multiply each side by 2 :

8 inches = base

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Can you solve this for me

jenyasd209 [6]

The number in the blank box should be 44 because it said that "3 times x and 4 times y is 44".

You need to multiply the first equation by 3:

3x + 3y = 39.

Then, subtract the first equation from the second equation:

3x - 3x = 0

4y - 3y = y

44 - 39 = 5

Your new equation should look like this: y = 5

Now, substitute y = 5 into the first original equation:

x + 5 = 13

x = 8

Ans. (x, y) = (8, 5)

4 0

2 years ago

Help! Quick please!

Vesnalui [34]

B..............
c>=3 and n<0 and p<=(9-3 c)/n

8 0

3 years ago

Read 2 more answers

In this picture, m∠AOC = 65° and m∠COD = (2x + 4)°. If ∠AOC and ∠COD are complementary angles, then what is the value of x?

dmitriy555 [2]

Answer:

I think it's C hope it helps you

4 0

2 years ago

What is the distance in units between (−11, −20) and (−11, 5)

Morgarella [4.7K]

Distance = √(x₂-x₁)² + (y₂-y₁)²
d = √(-11-(-11)) + (5-(-20))²
d = √0² + 25²
d = √625
d = 25

In short, Your Answer would be 25 units

Hope this helps!

6 0

3 years ago

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When do we use the addition and subtraction formulas for sine or cosine?

777dan777 [17]

When we need the cosine of the sum of two angles, we use the addition formula for cosine.

When we need the sine of the difference of two angles, we use the difference formula for sine.

Let's try one. Say we needed to know

$\cos 15^\circ
$

We're only expected to know the trig functions of 30/60/90 and 45/45/90 triangles. Fortunately, $15^\circ = 45^\circ - 30^\circ$ so

$\cos 15^\circ = \cos( 45^\circ - 30^\circ )$

$= \cos( 45^\circ) \cos(30^\circ ) + \sin(45^\circ) \sin(30^\circ)$

$= \dfrac{\sqrt{2}}{2} \cdot \dfrac{ \sqrt{3}}{2} + \dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2}$

$\cos 15^\circ = \frac{1}{4} (\sqrt{6} + \sqrt{2})$

4 0

3 years ago