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4 years ago

Rewrite this expression using the additive inverse property. 65 – 37

Mathematics

1 answer:

tankabanditka [31]4 years ago

5 0

Answer:

37 + 42

Step-by-step explanation:

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Mr. Ravenhead's car averages 23.75 miles per gallon of gasoline. Express this amount as a mixed number in simplest form

lisabon 2012 [21]

A mixed number consists of a whole number and a fraction.

23.7523 will be the whole number.
0.75 will be the fraction.

To convert 0.75 into a fraction, set it over 100 and simplify.

$\frac{75}{100} = \frac{75:25}{100:25} = \frac{3}{4}$

$23 \frac{3}{4}$

4 0

4 years ago

Simplify the expression so there is only one positive power for each base. (5^-2 x 4^-4)^-2

faltersainse [42]

$(a\cdot b)^n=a^n\cdot b^n\\(a^n)^m=a^{n\cdot m}\\\\(5^{-2}\cdot4^{-4})^{-2}=(5^{-2})^{-2}\cdot(4^{-4})^{-2}=5^{-2\cdot(-2)}\cdot4^{-4\cdot(-2)}\\\\=\boxed{5^4\cdot4^8}\to\fbox{A.}$

6 0

3 years ago

Read 2 more answers

Andrew pays $15 for a haircut . He leaves a 20% tip. What is the total amount andrew pays for the haircut, including the tip?

beks73 [17]

Answer:

$18.00

Step-by-step explanation:

Start by working out 10%, that's easy:

15 ÷ 10 = 1.5

That means a 10% tip would be $1.50, so a 20% tip would be twice that amount:

1.50 x 2 = 3

A 20% tip is $3.00, and added on to the original $15.00, the total is $18.00

7 0

3 years ago

Problem Page At a certain college, 49% of the students are female, and 19% of the students major in civil engineering. Furthermo

SIZIF [17.4K]

Answer with Step-by-step explanation:

Let

A=Students are female

B=Students major in civil engineering

The probability that a students are female=P(A)= $\frac{49}{100}=0.49$

The probability that a student major in civil engineering=P(B)= $\frac{19}{100}=0.19$

The probability that students both are female= $P(A\cap B)=\frac{12}{100}=0.12$

a.We have to find the probability that a random selected female student major in civil engineering.

We have to find $P(B/A)$

$P(B/A)=\frac{P(A\cap B)}{P(A)}$

$P(B/A)=\frac{0.12}{0.49}=0.24$

Hence, the probability that a randomly selected female student majors in civil engineering=0.24

b.We have to find the probability that a random selected civil engineering major is female.

We have to find $P(A/B)$

$P(A/B)=\frac{P(A\cap B)}{P(B)}=\frac{0.12}{0.19}=0.63$

Hence, the probability that a randomly selected civil engineering major is female=0.63

3 0

3 years ago

Solve.

Ksenya-84 [330]

Solve.

x² + 4x + 4 = 18

A x=−4±3√ 2

B x=2±3√ 2

C x=4±9√ 2

D x=−2±3√2

 $x^2 + 4x + 4 = 18 \\ \\ x^2+4x+4-18= 0 \\ \\ x^2+4x-14=0 \\ \\ x_1_y_2= \dfrac{-b\pm \sqrt{b^2-4ac} }{2a} \qquad a= 1\qquad b= 4\qquad c= -14 \\ \\ \\ x_1_y_2= \dfrac{-4\pm \sqrt{4^2-4(1)(-14)} }{2(1)} \\ \\ \\ x_1_y_2= \dfrac{-4\pm \sqrt{16-(-56)} }{2} \\ \\ \\ x_1_y_2= \dfrac{-4\pm \sqrt{72} }{2} \\ \\ \\ x_1= \dfrac{-4+ \sqrt{72} }{2} \qquad\qquad x_2= \dfrac{-4+ \sqrt{72} }{2}\\ \\ \\ x_1= \dfrac{-4+ \sqrt{2^3*3^2} }{2} \qquad\qquad x_2= \dfrac{-4- \sqrt{2^3*3^2} }{2}$

 $x_1= \dfrac{-4+2*3 \sqrt{2} }{2} \qquad\qquad x_2= \dfrac{-4- 2*3\sqrt{2} }{2} \\ \\ \\ x_1= \dfrac{-4+6 \sqrt{2} }{2} \qquad\qquad x_2= \dfrac{-4- 6\sqrt{2} }{2} \\ \\ \\ x_1= \dfrac{-4}{2} + \dfrac{6 \sqrt{2} }{2} \qquad\qquad x_2= \dfrac{-4}{2}- \dfrac{ 6\sqrt{2} }{2} \\ \\ \\ x_1= -2 + 3 \sqrt{2} \qquad\qquad\quad x_2= -2- 3\sqrt{2} \\ \\ \\ \boxed{x= -2\pm 3\sqrt{2} } \to D)$

7 0

4 years ago

Read 2 more answers