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3 years ago

Find the inverse of the function.1x) = -3x-4

Mathematics

1 answer:

Varvara68 [4.7K]3 years ago

3 0

Answer:

(4+1x)/-3=x

I actually don't understand about the question, but I'll try to answer

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Step-by-step explanation:

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3 years ago

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Leviafan [203]

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3 years ago

In △ABC, point M is the midpoint of AB , point D∈ AC so that AD:DC=2:5. If AABC=56 yd2, find ABMC, AAMD, and ACMD.

Komok [63]

Since point M is the midpoint of AB, then AM=MB.

Consider the area of the triangles ABC and BMC:

$A_{ABC}=\dfrac{1}{2}\cdot AB\cdot h_c=56\ yd^2,$

where $h_c$ is the height drawn from the vertex C to the side AB.

So, $AB\cdot h_c=112\ yd^2.$

Now

$A_{BMC}=\dfrac{1}{2}\cdot BM\cdot h_c=\dfrac{1}{2}\cdot \dfrac{AB}{2}\cdot h_c=\dfrac{1}{4}\cdot AB\cdot h_c=\dfrac{1}{4}\cdot 112=28\ yd^2.$

Also

$A_{AMC}=A_{ABC}-A_{BMC}=56-28=28\ yd^2.$

Now consider the area of the triangles AMD and CMD. Let $h_M$ be the height drawn from the point M to the side AC.

$A_{AMD}=\dfrac{1}{2}\cdot AD\cdot h_M=\dfrac{1}{2}\cdot \dfrac{2AC}{7}\cdot h_M=\dfrac{2}{7}\cdot \left(\dfrac{1}{2}\cdot AC\cdot h_M\right)=\dfrac{2}{7}\cdot A_{AMC}=\dfrac{2}{7}\cdot 28=8\ yd^2.$

Therefore,

$A_{MDC}=A_{AMC}-A_{AMD}=28-8=20\ yd^2.$

Answer: $A_{MBC}=28\ yd^2, A_{AMD}=8\ yd^2, A_{MDC}=20\ yd^2.$

5 0

3 years ago

Read 2 more answers

-) 10(20 - 4x) = 400

Nadusha1986 [10]

Answer:

10(20-4x)=400

200-40x=400

400-200=4x

x=5

4 0

4 years ago

Roduct.<br> 93 + (68 + 7)

Pani-rosa [81]

Answer:

Step-by-step explanation:

93 + (68 + 7)= 168

6 0

4 years ago

Read 2 more answers

Find the inverse of the function.1x) = -3x-4​

Find the inverse of the function.1x) = -3x-4