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3 years ago

Heeeeeeeeeeeeeeeeeeeeeeeeeeeeelelllllllllllllpppppppppppppppp

Mathematics

1 answer:

Over [174]3 years ago

4 0

The scale factor is 1

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. Multiply (3x+2)(4x-7)

aleksklad [387]

Answer:

12 $x^{2}$ + 29x - 14

Step-by-step explanation:

(3x+2)(4x-7)

Multiply out 3x first then 2

(3x * 4x) - (3x * -7) + (2 * 4x) + (2 * -7)

12 $x^{2}$ - (- 21x) + 8x + ( -14)

12 $x^{2}$ + 21x + 8x - 14 Combine like terms

12 $x^{2}$ + 29x - 14

4 0

3 years ago

Read 2 more answers

I’ll give Brainlest<br><br> m m m

JulijaS [17]

Answer:

steps below

Step-by-step explanation:

arc QR = 2x57 - 41 = 73

arc PQ = 360 - 73 - 41 - 137 = 109

m∠Q = (41+137) / 2 = 89°

m∠R = (109+137) / 2 = 123°

m∠S = (109+73) / 2 = 91°

check: ∠R+∠p = 123+57 = 180

∠Q+∠S = 89+91 = 180

3 0

3 years ago

A cookie recipe that yields 24 cookies requires 1 3/4 cups of butter. When the ingredients in this recipe are increased proporti

erik [133]

Answer:

5 1/4

Step-by-step explanation:

* is multiplication

1 3/4 is 1.75

24/1.75 = 72/×

1.75 * 72 = 24 * x

126 = 24x

24x = 126

x = 5.25 or 5 1/4

8 0

3 years ago

Simplify the expression (-8)^-3x15^0

avanturin [10]

The answer is -1 over 512 (fraction form)
ANSWER: -1/512

5 0

3 years ago

Find the volume of the following solid. The solid between the cylinder f(x,y)equals=e Superscript negative xe−x and the region

PilotLPTM [1.2K]

It is hard to comprehend your question. As far as I understand:

f(x,y) = e^(-x)

Find the volume over region R = {(x,y): 0<=x<=ln(6), -6<=y <= 6}.

That is all I understood. It would be easier to understand with a picture or some kind of visual aid.

Anyways, to find the volume between the surface and your rectangular region R, we must evaluate a double integral of f on the region R.

$\iint_{R}^{ } e^{-x}dA=\int_{-6}^{6} \int_{0}^{ln(6)} e^{-x}dx dy$

Now evaluate,

$\int_{0}^{ln(6)}e^{-x}dx$

which evaluates to, 5/6 if I did the math correct. Correct me if I am wrong.

Now integrate this w.r.t. y:

$\int_{-6}^{6}\frac{5}{6}dy = 10$

So,

$\iint_{R}^{ } e^{-x}dA=\int_{-6}^{6} \int_{0}^{ln(6)} e^{-x}dx dy = 10$

7 0

3 years ago