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4 years ago

Help me image in side worth 99 points!

Mathematics

2 answers:

Bas_tet [7]4 years ago

8 0

The total is 90°. Set the equation

8x - 6 + 80 = 90

Simplify. Combine like terms

8x + 74 = 90

Isolate the x. Do the opposite of PEMDAS. What you do to one side, you do to the other

8x + 74 (-74) = 90 (-74)

8x = 16

8x/8 = 16/8

x = 2

2 is your answer for x

hope this helps

mamaluj [8]4 years ago

4 0

You take (8x - 6) then add 80 =90 then solve for x
8x-6+80 =90
x= 2

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The value of a is 63.43 degrees

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3 years ago

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Rationalize the denominator in the expression. <br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B4%7D%7B%5Csqrt%7B2%7D%20%7D%20

Dovator [93]

Answer:

2 $\sqrt{2}$

Step-by-step explanation:

See the photo for the steps. :)

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3 years ago

2.5 ml medication.how many teaspoon should be given

Verdich [7]

5ml=1 teaspoon

2.5ml=1/2 teaspoon

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3 years ago

Will give brainlyist

Luba_88 [7]

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The 2nd one

Step-by-step explanation:

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8 0

3 years ago

A 10-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Alex777 [14]

Answer:

The area is changing at 15.75 square feet per second.

Step-by-step explanation:

The triangle between the wall, the ground, and the ladder has the following dimensions:

H: is the length of the ladder (hypotenuse) = 10 ft

B: is the distance between the wall and the ladder (base) = 6 ft

L: the length of the wall (height of the triangle) =?

dB/dt = is the variation of the base of the triangle = 9 ft/s

First, we need to find the other side of the triangle:

$H^{2} = B^{2} + L^{2}$

$L = \sqrt{H^{2} - B^{2}} = \sqrt{(10)^{2} - B^{2}} = \sqrt{100 - B^{2}}$

Now, the area (A) of the triangle is:

$A = \frac{BL}{2}$

Hence, the rate of change of the area is given by:

$\frac{dA}{dt} = \frac{1}{2}[L*\frac{dB}{dt} + B\frac{dL}{dt}]$

$\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - B^{2}}*\frac{dB}{dt} + B\frac{d(\sqrt{100 - B^{2}})}{dt}]$

$\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - B^{2}}*\frac{dB}{dt} - \frac{B^{2}}{(\sqrt{100 - B^{2}})}*\frac{dB}{dt}]$

$\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - 6^{2}}*9 - \frac{6^{2}}{\sqrt{100 - 6^{2}}}*9]$

$\frac{dA}{dt} = 15.75 ft^{2}/s$

Therefore, the area is changing at 15.75 square feet per second.

I hope it helps you!

5 0

3 years ago