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3 years ago

Callie reads 5 pages in 20 minutes. She spends the same amount of time per page.

Mathematics

1 answer:

Ira Lisetskai [31]3 years ago

8 0

Answer:

Step-by-step explanation:

You said it

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If f(x) = 2x² - 6x + 8, find f(a - 2).

yan [13]

Answer: f(a-2) = 2a² - 14a + 28

Step-by-step explanation:

f(x) = 2x² - 6x + 8

you can take a-2 and substitute it for each x, which would look like this:

f(a-2) = 2(a-2)² - 6(a-2) + 8

then distribute those in ( )

f(a-2) = 2a² - 8a + 8 - 6a + 12 + 8

combine like terms

f(a-2) = 2a² - 14a + 28

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Your _______ should furnish enough money to live on, in an emergency, for six months

OLEGan [10]

Your bank should furnish enough money to live on , in an emergency , for six months.
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3 years ago

Use the diagram to solve for x.

Andreyy89

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3 years ago

Evaluate the function. g(x) = 3x^2 – 2x + 5 Find f(2)

Nadya [2.5K]

Answer:

Step-by-step explanation:

Replace X with 2

Evaluate the function. g(x) = 3x^2 – 2x + 5 Find f(2)

g(x) = 3(2)^2 – 2(2) + 5

Next conduct PEMDAS

Exponents are first so solve 2^2 which is 2 x 2 = 4

g(x) = 3(4) – 2(2) + 5

Next step is multiplication multiply 3 x 4 and 2x2

g(x) = 12 – 4 + 5

conduct adding and subtracting left from right

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2 years ago

Find the square roots of each complex number. a. 5 + 12i b. 5 − 12i

musickatia [10]

ANSWER:

Square root of 5 + 12i is 3+2i or -3 – 2i and square root of 5 – 12i is 3 – 2i or -3 + 2i.

SOLUTION:

Given, we have to find square roots of 5 + 12i and 5 – 12i

a) 5 + 12i

Now, square root of 5 + 12i

Suppose that a + bi is a square root of 5 + 12i.

$\text { Then, }(a+b i)^{2}=\left(a^{2}-b^{2}\right)+(2 a b) i=5+12 i$

Equate real and imaginary parts:

$\begin{array}{l}{a^{2}-b^{2}=5 \text { and } 2 a b=12 \rightarrow b=\frac{6}{a}} \\\\ {\text { So, } a^{2}-\left(\frac{6}{a}\right)^{2}=5} \\\\ {\rightarrow a^{2}-\frac{36}{a^{2}}=5} \\\\ {\rightarrow a^{4}-5 a^{2}-36=0} \\\\ {\rightarrow\left(a^{2}-9\right)\left(a^{2}+4\right)=0}\end{array}$

Since a must be real, a = 3 or -3.

This gives b = 2 or -2, respectively.

Thus, we have two square roots: 3+2i or -3-2i.

b) 5 - 12i

Now, square root of 5 - 12i

Suppose that a - bi is a square root of 5 - 12i.

$\text { Then, }(a-b i)^{2}=\left(a^{2}-b^{2}\right)-(2 a b) i=5-12 i$

Equate real and imaginary parts:

$\begin{array}{l}{a^{2}-b^{2}=5 \text { and } 2 a b=12 \rightarrow b=\frac{6}{a}} \\\\ {S o, a^{2}-\left(\frac{6}{a}\right)^{2}=5} \\\\ {\rightarrow a^{2}-\frac{36}{a^{2}}=5} \\\\ {\rightarrow a^{4}-5 a^{2}-36=0} \\\\ {\rightarrow\left(a^{2}-9\right)\left(a^{2}+4\right)=0} \\\\ {\text { since a must be real, } a=3 \text { or }-3}\end{array}$

This gives b = 2 or -2, respectively.

Thus, we have two square roots: 3 - 2i or -3 + 2i.

Hence, square root of 5 + 12i is 3+2i or -3 – 2i and square root of 5 – 12i is 3 – 2i or -3 + 2i.

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3 years ago