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3 years ago

What is the product?

Mathematics

2 answers:

RSB [31]3 years ago

4 0

Answer:

15a^5b^15

Step-by-step explanation:

you multiply 5 and 3 to get 15, and then when you are multiplying exponents of variables, you actually add them, so a^2 times a^3 is actually a^5, and b^7 times b^8 is b^15. Because these are all multiplied together, you get the answer of 15a^5b^15.

sukhopar [10]3 years ago

4 0

Answer:

15 a^5 b^15

Step-by-step explanation:

3 a^2 b^7 * 5a^3 b^8

We know x^y * x^z = x^(y+z)

3*5 a^(2 +3) b^(7+8)

15 a^5 b^15

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4.5.36

Pavel [41]

Answer:

Step-by-step explanation:

Given function is h(t) = -16t² + 1500

a). For h(t) = 1000 feet,

1000 = -16t² + 1500

1000 - 1500 = -16t² + 1500 - 1500

-500 = -16t²

t² = $\frac{500}{16}$

t = $\sqrt{31.25}$

t = 5.59 sec

b). For h(t) = 940 feet,

940 = -16t² + 1500

940 - 1500 = -16t² + 1500 - 1500

-16t² = -560

t² = $\frac{-560}{-16}$

t = $\sqrt{35}$

t = 5.92 sec

c). For domain and range of the function,

When the jumper comes down to the ground,

h = 0

0 =-16t² + 1500

t² = $\frac{1500}{16}$

t = $\sqrt{93.75}$

t = 9.68 sec

Since, x-values on the graph vary from x = 0 to x = 9.68,

Domain : [0, 9.68]

Vertex of the quadratic function: (0, 1500)

Since, coefficient of the highest degree term is negative, parabola will open downwards.

Therefore, y-values of the function will vary in the interval y = 0 to y = 1500

Range: [0, 1500]

3 0

3 years ago

Functions. Type the correct domain in the box

Mnenie [13.5K]

Function:

y = -2/3 x + 7

Substitute the given values to their respective variable and solve.

x y

-6 3

3 5

15 -3

-12 15

7 0

3 years ago

How do you this question (pre-cal)

Novay_Z [31]

The goal to proving identities is to transform one side into the other. We can only pick one side to transform while the other side stays the same the entire time. The general rule of thumb is to transform the more complicated side (though there may be exceptions to this guideline).

So I'll take the left hand side and try to turn it into $\csc^2( B )$

One way we can do that is through the following steps:

$\frac{\tan(B) + \cot(B)}{\tan(B)} = \csc^2(B)\\\\\frac{\tan(B)}{\tan(B)} + \frac{\cot(B)}{\tan(B)} = \csc^2(B)\\\\1 + \cot(B)*\frac{1}{\tan(B)} = \csc^2(B)\\\\1 + \cot(B)*\cot(B) = \csc^2(B)\\\\1 + \cot^2(B) = \csc^2(B)\\\\1 + \frac{cos^2(B)}{\sin^2(B)} = \csc^2(B)\\\\\frac{sin^2(B)}{\sin^2(B)}+\frac{cos^2(B)}{\sin^2(B)} = \csc^2(B)\\\\\frac{sin^2(B)+cos^2(B)}{\sin^2(B)} = \csc^2(B)\\\\\frac{1}{\sin^2(B)} = \csc^2(B)\\\\\csc^2(B)=\csc^2(B) \ \ {\Large \checkmark}\\\\$