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3 years ago

Please help me with this 6 problem please please and I need work to please

Mathematics

1 answer:

klio [65]3 years ago

8 0

What do you need help with? it helps us to know the question!

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I need help! what is 10*80/9+6r equal

S_A_V [24]

94r8it the answer can I please have brainelist

3 0

3 years ago

Which statement is true? 1, A strong association implies a positive correlation. 2, A Positive correlation implies a strong asso

Gnesinka [82]

Answer:

it's A

Step-by-step explanation:

is a positive correlation

8 0

2 years ago

Tiffany used to live 15 kilometers away from the school, but after she moved she now lives 9.9 kilometers away. She is _____ per

valkas [14]

Answer:

Tiffany is 34% closer to the school.

Step-by-step explanation:

Given the statement: Tiffany used to live 15 kilometers away from the school, but after she moved she now lives 9.9 kilometers away.

She lived away from the school = 15 km

as, she moved 9.9 km away.

Now, the distance closer to school from where she lived = 15 -9.9 = 5.1 km

To, find how much percent she is closer to school.

Percent states that a number or ratio expressed as a fraction of 100.

$percent = \frac{5.1}{15} \times 100$ = $\frac{51 \times 100}{150} = \frac{51 \times 10}{15}$

Simplify:

percent closer to school = 34%

Therefore, she is 34% closer to the school.

5 0

3 years ago

Please help <br> I will mark brainlist

Naya [18.7K]

Answer:

B x = sqrt(48)

Step-by-step explanation:

Half of the bottom side measures 4.

4^2 + x^2 = 8^2

16 + x^2 = 64

x^2 = 48

x = sqrt(48)

Answer: B x = sqrt(48)

4 0

2 years ago

Simplify f+g / f-g when f(x)= x-4 / x+9 and g(x)= x-9 / x+4

steposvetlana [31]

$f(x)=\dfrac{x-4}{x+9};\ g(x)=\dfrac{x-9}{x+4}\\\\f(x)+g(x)=\dfrac{x-4}{x+9}+\dfrac{x-9}{x+4}=\dfrac{(x-4)(x+4)+(x-9)(x+9)}{(x+9)(x+4)}\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{x^2-4^2+x^2-9^2}{(x+9)(x+4)}=\dfrac{2x^2-16-81}{(x+9)(x+4)}=\dfrac{2x^2-97}{(x+9)(x+4)}\\\\f(x)-g(x)=\dfrac{x-4}{x+9}-\dfrac{x-9}{x+4}=\dfrac{(x-4)(x+4)-(x-9)(x+9)}{(x+9)(x+4)}\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{x^2-4^2-(x^2-9^2)}{(x+9)(x+4)}=\dfrac{x^2-16-x^2+81}{(x+9)(x+4)}=\dfrac{65}{(x+9)(x+4)}$

$\dfrac{f+g}{f-g}=(f+g):(f-g)=\dfrac{2x^2-97}{(x+9)(x+4)}:\dfrac{65}{(x+9)(x+4)}\\\\=\dfrac{2x^2-97}{(x+9)(x+4)}\cdot\dfrac{(x+9)(x+4)}{65}\\\\Answer:\ \boxed{\dfrac{f+g}{f-g}=\dfrac{2x^2-97}{65}}$

6 0

3 years ago

Read 2 more answers