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2 years ago

U = 2x - 2 (Solve for x) Hurry!

Mathematics

1 answer:

kobusy [5.1K]2 years ago

4 0

X=1

because you would add two to the other side and have 2x=2 divide by 2 and you would be left with x=1

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Apply the distributive property to write an equivalent expression 4(2+w)

Alja [10]

Hi 4w+8 is the answer

8 0

3 years ago

The parabola y = x² - 4 opens: A.) up B.) down C.) right D.) left

fomenos

Answer:

Step-by-step explanation:

Here the easy rules to remember the orientation of the parabolas are

a) If x is squared it opens up or down. And its coefficient of { $x^{2}[tex] is negative it opens down.b) If y is squared it opens side ways right or left. It its coefficient of [tex]y^{2}$

Hence in our equation of parabola

$y = x^ 2-4$

x is squared and its coefficient is positive , hence it opens up towards positive y axis.

3 0

2 years ago

URGENT NOW: Use the graph below to fill in the blank with the correct number: f(−2) = _______ X, Y graph. Plotted points negativ

navik [9.2K]

The answer would be 2. As you can see on the point, you already have (-2,2) as a point. F(x) equals y in a sense. So when plugging in x into a function, you should get its respect y value. In the function graph, -2 yields 2. Therefore, your answer is 2.

6 0

3 years ago

What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove

wolverine [178]

Answer:

The solutions are:

$b=0,\:b=4$

Step-by-step explanation:

Considering the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

Solving the expression

$\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}$

$\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c$

$5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2$

$5b^3-10=6b^3-4b^2-10$

$\mathrm{Switch\:sides}$

$6b^3-4b^2-10=5b^3-10$

$6b^3-4b^2-10+10=5b^3-10+10$

$6b^3-4b^2=5b^3$

$\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}$

$6b^3-4b^2-5b^3=5b^3-5b^3$

$b^3-4b^2=0$

$Using\:the\:Zero\:Factor\:Principle:$ $if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$