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3 years ago

ALGEBRA 1:

Mathematics

2 answers:

Fed [463]3 years ago

4 0

Answer:

x^2 - 16x + 64

Step-by-step explanation:

(x-8)^2

(x-8) (x-8)

x(x-8) -8(x-8)

x*x+x*-8-8x-8*-8

x^2-8x-8x+16

x^2 - 16x + 64

Phantasy [73]3 years ago

3 0

64-16x+x^2 i had this same question and that’s the answer i put and for it right :)

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What is 2000 x 5000 in long mupulitycation form

Studentka2010 [4]

Answer:

10,000,000

Step-by-step explanation:

You multiply all of the digits on the bottom starting from right to left with the numbers on the top (Biggest number on top smallest on bottom). after you multiplied 5000 by the first zero in 2000 you go down a line (Lined paper is the best to use) and add a zero as a place holder and then multiply 5000 by the second zero in 2000 then after that you go down a nother line and add two zeros for a place holder... you repeat the steps of multiplying and go down another line and add three zeros as place holders. after your finnished multiplying you add all the numbers in your lines together and you will get 10,000,000

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4 years ago

bob is picking randomly from a bag containing tiles member 1 to 10. write down the probability that the number he picks is 7

KengaRu [80]

Answer:

Step-by-step explanation:

a) No of outcomes= 7

total no of outcomes= 10

probability = no of outcomes/total no of outcomes

=7/10

b) 4/10 ( since the outcomes could be 1,2,3,4)

c) no 1 to 10 has 5 odd numbers

so, probability = 5/10 = 1/2

d) multiples of 3 from 1 to 10 are :-

3,6,9= 3 no's

so probability = 3/10

4 0

3 years ago

I just need some help solving this question, i’m not sure what to do

KonstantinChe [14]

From the double-angle identity,

$cos2x=2*sinx*cosx$

we can rewritte our given equation as:

$4sinxcosx-2cosx=0$

By factoring 2cosx on the left hand side, we have

$2cosx(2sinx-1)=0$

This equation has 2 solutions when

$\begin{gathered} cosx=0\text{ ...\lparen A\rparen} \\ and \\ 2sinx-1=0\text{ ...\lparen B\rparen} \end{gathered}$

From equation (A), we obtain

$x=\frac{\pi}{2}\text{ or }\frac{3\pi}{2}$

and from equation (B), we have

$\begin{gathered} sinx=\frac{1}{2} \\ which\text{ gives} \\ x=\frac{\pi}{6}\text{ or }\frac{5\pi}{6} \end{gathered}$

On the other hand, we can find one more solution from the original equation by substituting x=0, that is,

$\begin{gathered} 2ccos(2\times0)-2cos0=0 \\ which\text{ gives} \\ 2\times1-2\times1=0 \\ so\text{ 0=0} \end{gathered}$

then, x=0 is another solution. In summary, we have obtained the following solutions:

$\begin{gathered} x=0 \\ x=\frac{\pi}{2}\text{or}\frac{3\pi}{2}\text{ and } \\ x=\frac{\pi}{6}\text{or}\frac{5\pi}{6} \end{gathered}$

However, the intersection of the last set is empty. So the unique solution is x=0 as we can corroborate on the following picture:

Therefore, the solution set is: {0}

7 0

1 year ago

Jason is buying a car, and he wants to pay for it in 48 monthly installments. If the total cost of the car is $9,300, how much w

Dmitrij [34]

The answer is 193.75 per month

3 0

3 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