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3 years ago

I need help on number 1

Mathematics

2 answers:

ollegr [7]3 years ago

7 0

2.5. is the answer. ..

Mashutka [201]3 years ago

6 0

What you do is divide those two numbers. what is 80 divided by 32? the answer is 2.5

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A number b times -16 is greater than 5

lord [1]

Write the Inequality

A number b times -16 is greater than 5.

A number b is a variable... b.

Times means multiply.

-16 is what you are multiplying b by.

Is greater than means the greater than symbol.

5 is what the past statements are greater than.

So this is the inequality.

-16b > 5

Solve the Inequality

You need to isolate b.

To do this, divide each side by -16. You do this because b is being multiplied by -16. You know that any number divided by that number equals 1, canceling the number.

-16b ÷ -16 = b

5 ÷ -16 = -5/16

So you get b > -5/16

So to make the inequality true, b must be more than -5/16.

Or as an inequality...

b > -5/16

8 0

2 years ago

Hello! :)

stepan [7]

Answer:

Step-by-step explanation:

7 0

2 years ago

Anwser Please unit test!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

madam [21]

The answer is B, d=t+100. (Can I have Brainlest?)

8 0

3 years ago

Read 2 more answers

B) there is a line with positive slope that is tangent to both circles. find the equation of this line, and (c) determine the po

IgorC [24]

Given that there is a line with positive slope that is tangent to both circles $x^2 + y^2 = 1$ and $(x-3)^2 + y^2 = 4$ .

Let the point of tangency in the first circle be (a, b) and the point of tangency in the second circle be (c, d), then the slope of the tangent to the first circle is given by

$2x+\left.2y \frac{dy}{dx}\right|_{(a, b)} =0 \\ \\ \Rightarrow \left.\frac{dy}{dx}\right|_{(a, b)} =- \frac{x}{y} =- \frac{a}{b}$

and the slope of the second circle is given by

$2(x-3)+\left.2y \frac{dy}{dx}\right|_{(c, d)} =0 \\ \\ \Rightarrow \left.\frac{dy}{dx}\right|_{(a, b)} =- \frac{x-3}{y} =- \frac{c-3}{d}$

From the first circle equation, we have

$x^2 = 1 - y^2 \\ \\ \Rightarrow a^2 = 1 - b^2$

and from the second equation, we have

 $(x-3)^2 = 4 - y^2 \\ \\ (c-3)^2 = 4 - d^2$ 

Since the two slopes above are for the same line, then

 $\frac{a}{b}= \frac{c-3}{d} \\ \\ \Rightarrow ad=b(c-3) \\ \\ \Rightarrow a^2d^2=b^2(c-3)^2 \\ \\ \Rightarrow(1-b^2)d^2=b^2(4-d^2) \\ \\ \Rightarrow d^2-b^2d^2=4b^2-b^2d^2 \\ \\ \Rightarrow d^2=4b^2 \\ \\ \Rightarrow d=\pm2b$

Thus, we have

$\frac{a}{b}= \frac{c-3}{2b} \\ \\ \Rightarrow a= \frac{c-3}{2}$

7 0

2 years ago

Ryan and expression to represent the perimeter of a subtract 3 to 8 and three a plus one

Andrew [12]

9
subtract,3-8 = 5
add,5+3+1 =9

4 0

3 years ago