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kondor19780726 [428]
3 years ago
13

Help please !!!!! i dont know how

Mathematics
1 answer:
dem82 [27]3 years ago
8 0

hello since we are simplifying 15^-3*15^6 all we have to do is add the exponents when we add them we end up with an answer of 15^3

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Please help me please I need help!!!!!!!!!!!!!!!!!!!hshdjdndndbdbnfjfkfkfkdkdkfkfjdjfiid
Serggg [28]

y = (x - 2)(x + 5)    First I would multiply (x - 2) and (x + 5) together

y = x² + 5x - 2x - 10

y = x² + 3x - 10


Next I would plug in a number for x to find its y value. I will plug in 0

y = 0² + 3(0) - 10

y = -10

(0, -10)   [this is the y-intercept (the y value when x = 0)]


Your answer is Graph A because the y-intercept is (0,-10)

6 0
3 years ago
Help please<br><br>What is 2/3÷1/2​
Luba_88 [7]

Answer: B

Step-by-step explanation:

2/3 x 2/1 =4/3

3 0
2 years ago
If y varies inversely as x and y=16 when x=4, find y when x=3
iren [92.7K]

Answer:

21.33 to nearest hundredth

Step-by-step explanation:

Inverse Variation is y = k / x where k is a constant.

Plug in the given values to find k:-

16 = k/4

k = 4*16 = 64 so the equation  of variation is y = 64/x.

So when x = 3,   y = 64/3 = = 21.33 to nearest hundredth.

6 0
3 years ago
What is the distance between (2,-1) and (2,5) rounded to the nearest 10th
xxMikexx [17]

Answer:

<h3>             5.0</h3>

Step-by-step explanation:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\d=\sqrt{(2-2)^2+(5+1)^2}=\sqrt{0^2+5^2}=\sqrt{25}=5

5 0
3 years ago
Integration of ∫(cos3x+3sinx)dx ​
Murljashka [212]

Answer:

\boxed{\pink{\tt I =  \dfrac{1}{3}sin(3x)  - 3cos(x) + C}}

Step-by-step explanation:

We need to integrate the given expression. Let I be the answer .

\implies\displaystyle\sf I = \int (cos(3x) + 3sin(x) )dx \\\\\implies\displaystyle I = \int cos(3x) + \int sin(x)\  dx

  • Let u = 3x , then du = 3dx . Henceforth 1/3 du = dx .
  • Now , Rewrite using du and u .

\implies\displaystyle\sf I = \int cos\ u \dfrac{1}{3}du + \int 3sin \ x \ dx \\\\\implies\displaystyle \sf I = \int \dfrac{cos\ u}{3} du + \int 3sin\ x \ dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3}\int \dfrac{cos(u)}{3} + \int 3sin(x) dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3} sin(u) + C +\int 3sin(x) dx \\\\\implies\displaystyle \sf I = \dfrac{1}{3}sin(u) + C + 3\int sin(x) \ dx \\\\\implies\displaystyle\sf I =  \dfrac{1}{3}sin(u) + C + 3(-cos(x)+C) \\\\\implies \underset{\blue{\sf Required\ Answer }}{\underbrace{\boxed{\boxed{\displaystyle\red{\sf I =  \dfrac{1}{3}sin(3x)  - 3cos(x) + C }}}}}

6 0
3 years ago
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