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3 years ago

Find the distance between the points (6,-4) and (-3,-4) .

Mathematics

1 answer:

umka2103 [35]3 years ago

3 0

The right answer is 9 units.

please see the attached picture for full solution

Hope it helps

Good luck on your assignment:)

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I didnt put the answer choices because ppl be putting random stuff

lesya692 [45]

I’m pretty sure the answer is 127.48 ,but I could be wrong.

5 0

3 years ago

What is the center and spread of the data

liraira [26]

Sorry if i am wrong but it's 40

7 0

3 years ago

Show your full solution

alex41 [277]

Answer:

1.the square pan is larger by 12 inches

40 cm^2

Step-by-step explanation:

The difference in size can be determined by calculating the difference between the perimeter of the square pan and the circumference of a circle

perimeter of a square = 4 x length

14 x 4 = 56 inches

Circumference of a circle = πD = (22/7) x 14 = 44 inches

Difference in size = 56 - 44 = 12 inches

Area of a rectangle = length x width

10 x 4 - 40 cm^2

3 0

3 years ago

Aiden knows the length of only one side of his garden. He says he will be able to find the area knowing only one aide. Explain h

Snowcat [4.5K]

This is possible providing the garden is in the shape that has equal sides, such as a square or an equilateral triangle.

Area of a square is found by multiplying one side to the other, which would be the same as multiplying two same numbers.

6 0

3 years ago

Read 2 more answers

What is the range of the equation

a_sh-v [17]

The range of the equation is $y>2$

Explanation:

The given equation is $y=2(4)^{x+3}+2$

We need to determine the range of the equation.

<u>Range:</u>

The range of the function is the set of all dependent y - values for which the function is well defined.

Let us simplify the equation.

Thus, we have;

$y=2 \cdot 4^{x+3}+2$

This can be written as $y=2^{1+2(x+3)}+2$

Now, we shall determine the range.

Let us interchange the variables x and y.

Thus, we have;

$x=2^{1+2(y+3)}+2$

Solving for y, we get;

$x-2=2^{1+2(y+3)}$

Applying the log rule, if f(x) = g(x) then $\ln (f(x))=\ln (g(x))$ , then, we get;

$\ln \left(2^{1+2(y+3)}\right)=\ln (x-2)$

Simplifying, we get;

$(1+2(y+3)) \ln (2)=\ln (x-2)$

Dividing both sides by $\ln (2)$ , we have;

$2 y+7=\frac{\ln (x-2)}{\ln (2)}$

Subtracting 7 from both sides of the equation, we have;

$2 y=\frac{\ln (x-2)}{\ln (2)}-7$

Dividing both sides by 2, we get;

$y=\frac{\ln (x-2)-7 \ln (2)}{2 \ln (2)}$

Let us find the positive values for logs.

Thus, we have,;

$x-2>0$

$x>2$

The function domain is $x>2$

By combining the intervals, the range becomes $y>2$

Hence, the range of the equation is $y>2$

7 0

4 years ago