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2 years ago

Math question help pls.

Mathematics

1 answer:

devlian [24]2 years ago

6 0

Answer:

i dont know the first one, but the answer to the second question is 28

Step-by-step explanation:

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Given the point and it''s image, determine the scale factor. A (3, 6) A'' (4.5, 9) G'' (3, 6) G (1.5, 3) B (2, 5) B'' (1, 2.5).

Natasha2012 [34]

Put the value of the letters with " over the original
So for A we have 3 in the front and 4.5 for A"
4.5/3 = 1.5
We can check if this is actually the scale factor but using the other numbers
The second numbers for A and A" are 6 and 9
6 * 1.5 = 9
So for point A it is 1.5
Use the same thing for G. That gives you 2
For B, you get 0.5
So the scale factors are 1.5, 2, and 0.5

5 0

2 years ago

= The average score of Karl's first four math tests is 72 points. After the fifth and the sixth tests, his average score increas

mylen [45]

Answer:

83 points

Step-by-step explanation:

-The formula for average is given as:

$\bar X=\frac{1}{n}\sum{x_i}\\\\x_i=Individual \ Test \ scores\\n=number \ of \ scores$

Let a be his 5th score and (a+4) of the 6th

#We subtract the sum of the first 4 tests from the 6, to get the sum of the 5th and 6th tests:

$(6^{th}+5^{th})=\bar X n_6-\bar X n_4\\\\=75\times 6-72\times 4\\\\=162$

#We substitute for the 5th and 6th to solve a:

$5^{th}=a\\6^{th}=a+4\\\\a+(a+4)=162\\\\2a=158\\\\a=79\\\\6^{th}=a+4=79+4=83$

Hence, Karl score 83 points in his 6th test.

5 0

2 years ago

Please help me ahhh im doing 8th grade math with pythagorean theorem pleasr help me on number 10

otez555 [7]

The answer is 6.3 see pic

5 0

2 years ago

If x^3 + 1/x^3 = 110 then find x + 1/x

Svetlanka [38]

$Use:(a+b)^3=a^3+3a^2b+3ab^2+b^3\ (*)\\\\x^3+\left(\dfrac{1}{x}\right)^3\\=\underbrace{x^3+3\cdot x^2\cdot\dfrac{1}{x}+3\cdot x\cdot\left(\dfrac{1}{x}\right)^2+\left(\dfrac{1}{x}\right)^3}_{(*)}-3\cdot x^2\cdot\dfrac{1}{x}-3\cdot x\cdot\left(\dfrac{1}{x}\right)^2\\\\=\left(x+\dfrac{1}{x}\right)^3-3x-3\cdot\dfrac{1}{x}=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)$

therefore

$x^3+\left(\dfrac{1}{x}\right)^3=110\iff \left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)=110\\\\subtitute\ t=x+\dfrac{1}{x}\\\\t^3-3t=110\ \ \ \ |subtract\ 110\ from\ both\ sides\\\\t^3-3t-110=0\\\\t^3-25t+22t-110=0\\\\t(t^2-25)+22(t-5)=0\ \ \ |use\ a^2-b^2=(a-b)(a+b)\\\\t(t-5)(t+5)+22(t-5)=0\\\\(t-5)[t(t+5)+22]=0\iff t-5=0\ or\ \underbrace{t^2+5t+22=0}_{no\ solution}\\\\t=5\iff x+\dfrac{1}{x}=5\\\\Answer:\boxed{x+\dfrac{1}{x}=5}$

3 0

2 years ago

A park is a rectangle that is 33 miles wide and 56 miles long. How long is the diagonal?

loris [4]

that is the answer

good luck

and follow please

4 0

2 years ago