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4 years ago

7x = 90 +22 Solve this problem

Mathematics

1 answer:

prisoha [69]4 years ago

5 0

Answer: X=16

Step-by-step explanation: Add 90 and 22 and you will get 112. Since the beginning of the equation is "7x" which is multiplication you use Inverse Operations. So, you divide the sum, 112, by 7. Your answer will be 16.

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Alexia's bathroom has a tub in the shape of a rectangular prism with a length of 1.5 meters, a width of 0.5 meter, and a height

disa [49]

The answer would be 0.3m^3 . Good luck! :)

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4 years ago

Roberto needs 6 cookies and 2 brownies for every 4 plates he makes for a bake sale. Drag cookies and brownies into the box to sh

ch4aika [34]

6 cookies for every 4 plates....6/4 = 1.5 cookies per plate
2 brownies for 4 plates.....2/4 = 0.5 brownies per plate

for 10 plates...
1.5(10) = 15 cookies <==
0.5(10) = 5 brownies <==

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3 years ago

The complement of set M is ______.

Firlakuza [10]

Answer:

M' is {-5, -4, -3, -2, -1, 0, 1, 3, 5, 6}

Step-by-step explanation:

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3 years ago

The base of an isosceles triangle is 4/5. The perimeter of the triangle is 4\2\15 what is the length of either of the remaining

crimeas [40]

Answer:

1 2/3

Step-by-step explanation:

Assuming you mean the perimeter is 4 2/15, the sum of the lengths of the two equal sides is ...

4 2/15 - 4/5 = 4 2/15 - 12/15 = 4 -10/15

Then either of the remaining sides is half this value, or ...

(4 -10/15)/2 = (4/2) -(1/2)(10/15) = 2 -5/15 = 1 10/15 = 1 2/3

Each of the remaining sides has length 1 2/3.

7 0

3 years ago

P(k)=a^k=2 3 4 find value of a that makes this is a valid probability distribution

Vesna [10]

Sounds like you're asked to find $a$ such that

$\displaystyle\sum_{k=2}^4\mathbb P(k)=\mathbb P(2)+\mathbb P(3)+\mathbb P(4)=1$

In other words, find $a$ that satisfies

$a^2+a^3+a^4=1$

We can factorize this as

$a^4+a^3+a^2-1=a^3(a+1)+(a-1)(a+1)=(a+1)(a^3+a-1)=0$

In order that $\mathbb P(k)$ describes a probability distribution, require that $\mathbb P(k)\ge0$ for all $k$ , which means we can ignore the possibility of $a=-1$ .

Let $a=y+\dfrac xy$ .

$a^3+a-1=\left(y+\dfrac xy\right)^3+\left(y+\dfrac xy\right)-1=0$
$\left(y^3+3xy+\dfrac{3x^2}y+\dfrac{x^3}{y^3}\right)+\left(y+\dfrac xy\right)-1=0$

Multiply both sides by $y^3$ .

$y^6+3xy^4+3x^2y^2+x^3+y^4+xy^2-y^3=0$

We want to find $x\neq0$ that removes the quartic and quadratic terms from the equation, i.e.

$\begin{cases}3x+1=0\\3x^2+x=0\end{cases}\implies x=-\dfrac13$

so the cubic above transforms to

$y^6-y^3-\dfrac1{27}=0$

Substitute $y^3=z$ and we get

$z^2-z-\dfrac1{27}=0\implies z=\dfrac{9+\sqrt{93}}{18}$
$\implies y=\sqrt[3]{\dfrac{9+\sqrt{93}}{18}}$
$\implies a=\sqrt[3]{\dfrac{9+\sqrt{93}}{18}}-\dfrac13\sqrt[3]{\dfrac{18}{9+\sqrt{93}}}$

6 0

4 years ago