6xy – бу • (7

Help me find the area of this hexagon.

Alika [10]

Answer:

70

Step-by-step explanation:

6 0

4 years ago

Chiara has 26 coins that equal 34 cents. All the coins are pennies, p, and nickels, n. How many nickels and pennies does Chiara

wlad13 [49]

Answer:

Chiara has 2 penny coins and 24 nickel coins

Step-by-step explanation:

Let

p ----> the number of penny coins

n ---> the number of nickel coins

Remember that

$1\ penny=\$0.01$

$1\ nickel=\$0.05$

Chiara has 26 coins that equal 34 cents

so

$p+n=26$

isolate the variable x

$p=26-n$ ----> equation A

34 cents=$0.34

$0.01p+0.05n=0.34$

Multiply by 100 both sides

$p+5n=34$

isolate the variable x

$p=34-5n$ -----> equation B

Create a table to guess

assume different values of n and determine the value of p in equation A and equation B

The solution is when the value of p in the equation A must be equal to the value of p in equation B

1) For n=1

equation A

$p=26-1=25$

equation B

$p=34-5(1)=29$

$29\neq 25$

2) For n=2

equation A

$p=26-2=24$

equation B

$p=34-5(2)=24\\24=24$

therefore

The solution is (2,24)

Chiara has 2 penny coins and 24 nickel coins

3 0

3 years ago

Read 2 more answers

Solve by taking the square root: (x-1)^2=9<br> Step by step PLEASE HELP!!!

Solnce55 [7]

$Answer: \\ {(x -1 )}^{2} = 9 \\ \Leftrightarrow \sqrt{ {(x - 1)}^{2} } = \sqrt{9} \\ \Leftrightarrow |x - 1| = 3 \\ \Leftrightarrow x - 1 = \pm3 \\ \Leftrightarrow x = - 2 \: \vee \: x = 4$

8 0

3 years ago

A point is refected across the x-axis. The new point is (5,-3.5).What is the distance between the two points?

Vesna [10]

When a point is reflected across the x-axis, it's y value is flipped. Here is a formula for your future reference:

X-axis reflection: (x,-y)

Using your given coordinates, you simply plug the numbers in, resulting in final coordinates of (5,3.5).

Because the problem asks for the distance between the two points, you now have to find the difference between -3.5 and 3.5, which is 7.

There you go; the distance between the points is 7 units.

5 0

4 years ago

Please check my homework on Negative Indices!

rusak2 [61]

Our first expression is $(2p^{-3} q^{8} )^{4}$ . Upon distributing the exponent 4 on all the terms, we get:

$(2p^{-3} q^{8} )^{4}=2^{4}(p^{-3})^{4}(q^{8})^{4}=16p^{-12}q^{32}=\frac{16q^{32}}{p^{12}}$

Therefore, your answer is correct for this part. :)

Second expression is $(2m^{-4}n^{5})^{-2}$ . Upon distributing the exponent -2 on all the terms, we get:

$(2m^{-4}n^{5})^{-2}=2^{-2}(m^{-4})^{-2}(n^{5})^{-2}=2^{-2}m^{8}n^{-10}=\frac{m^{8}}{4n^{10}}$

Your second answer is correct too.

Our third expression is $2(5g^{-4}h^{-6})^{2}$ . Upon distributing the exponent 2 on all the terms, we get:

$2(5g^{-4}h^{-6})^{2}=2(5^{2})(g^{-4})^{2}(h^{-6})^{2}=2(25)g^{-8}h^{-12}=\frac{50}{g^{8}h^{12}}$

This one is not correct. Your answer would have been correct, if the exponent were -2 instead of 2 in this part.

Our forth and last expression is $4(2c^{-3}d^{6})^{-5}$ . Upon distributing the exponent -5 on all the terms inside the parenthesis, we get:

$4(2c^{-3}d^{6})^{-5}=4(2^{-5})(c^{-3})^{-5}(d^{6})^{-5}=\frac{4}{2^{5}}(c^{15})(d^{-30})=\frac{4c^{15}}{32d^{30}}=\frac{c^{15}}{8d^{30}}$

Therefore, your answer for this part is also correct.

Looking at your work, I don't think you made a mistake in number 3 also, probably mis-typed the question while writing here :)

3 0

4 years ago

6xy – бу • (7 – 2х)