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3 years ago

What two two-digit numbers add to 48?

Mathematics

2 answers:

Ira Lisetskai [31]3 years ago

3 0

12 and 36 12+36 = 48

juin [17]3 years ago

3 0

21+27
40+08
37+11

Any two numbers that have are over 9 that add up to 48

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Mariel brought 48 brownies to school for a class party. The class ate 5/6 of the brownies, and Mariel brought the rest home. How

OverLord2011 [107]

Answer:

8 brownies

Step-by-step explanation:

Let's figure out how many brownies the class ate

Multiply 48* 5/6=40

Then, subtract how many the class ate from the total

48-40=8

So, he brought home 8 brownies

Hope this helped!

6 0

2 years ago

Read 2 more answers

Simplify the imaginary number sqr -75

m_a_m_a [10]

Answer:

5i $\sqrt{3}$

Step-by-step explanation:

Using the rule of radicals

$\sqrt{a}$ × $\sqrt{b}$ ⇔ $\sqrt{ab}$

and $\sqrt{-1}$ = i

Given

$\sqrt{-75}$

= $\sqrt{25(3)(-1)}$

= $\sqrt{25}$ × $\sqrt{3}$ × $\sqrt{-1}$

= 5 × $\sqrt{3}$ × i

= 5i $\sqrt{3}$

3 0

3 years ago

Is (x+7) a factor of f(x) x3-3x2+ 2x-8?

Snowcat [4.5K]

Answer:

(x+7) is not a factor

Step-by-step explanation:

we know that

If (x+7) is a factor of f(x)

then

For $x=-7, f(-7)=0$

Verify

we have

$f(x)=x^{3}-3x^{2} +2x-8$

Substitute x=-7

$f(-7)=(-7)^{3}-3(-7)^{2} +2(-7)-8$

$f(-7)=-343-147 -14-8$

$f(-7)=--512$

$-512\neq 0$

therefore

(x+7) is not a factor

3 0

2 years ago

Points RR, LL, and SS are:

vovikov84 [41]

Answer: C

Step-by-step explanation:

8 0

3 years ago

Divide 12x 4 y 4 by 3x 2 y. A. 4x 6y 5 B. 4x 2y 3 C. 9x 6 D. 9x 2y 3

andrew-mc [135]

Answer:

$\frac{12x^4y^4}{3x^2y}=4x^2y^3$

Step-by-step explanation:

Consider an element $a\neq 0$

We know that $\frac{a^m}{a^n}=a^{m-n}$

Here, m and n are exponents and a is a base. Exponent refers to the no. of times a term is multiplied by itself. Another word for exponents is powers.

Other Properties of exponents:

$a^m\times a^n=a^{m+n}\\a^0=1\\\left ( a^m \right )^n=a^{mn}$

Here, we are required to divide $12x^4y^4$ by $3x^2y$ .

Consider $\frac{12x^4y^4}{3x^2y}$

We will solve $\frac{12}{3}\,,\,\frac{x^4}{x^2}\,,\,\frac{y^4}{y}$ separately.

Here,

$\frac{12}{3}=4\\\frac{x^4}{x^2}=x^{4-2}=x^2\\\frac{y^4}{y}=y^3$

So, $\frac{12x^4y^4}{3x^2y}=4x^2y^3$

8 0

2 years ago