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3 years ago

Which equation matches the graph shown?

Mathematics

2 answers:

timofeeve [1]3 years ago

7 0

Take 2 points

(0,-3)
(1,1)

Slope:-

$\\ \sf\longmapsto m=\dfrac{1+3}{1}=4$

Equation of line in point slope form

$\\ \sf\longmapsto y-y_1=m(x-x_1)$

$\\ \sf\longmapsto y+3=4x$

$\\ \sf\longmapsto y=4x-3$

pogonyaev3 years ago

4 0

Answer:

C. y = 4x - 3

Step-by-step explanation:

The line has a positive slope and negative y-intercept.

m = 4, b = - 3

This is only matched by a choice C

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Find the exact value of csc ø, if cos ø= 1/9; 180° less ø less than 270°

Iteru [2.4K]

Answer:

The answer to your question is: the first option - 9/ $\sqrt{80}$

Step-by-step explanation:

Process

cosФ = adjacent side / hypotenuse

-1/9 = adjacent side / hypotenuse

adjacent side = -1 hypotenuse = 9

c² = a² + b²

b² = c² - a²

b² = 81 - 1

b = $\sqrt{80} \\$ b = opposite side

But b is negative because we are in the third quadrant.

b = - $\sqrt{80}$

Finally csc Ф = hypotenuse / opposite side

csc Ф = - 9 / $\sqrt{80}$

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3 years ago

X+2(x+y)=7 3(x-y)=1-2y What is x&y???? Please answer asap!!!!

Shkiper50 [21]

X= -x over 2 (negative x as a fraction and 2 is the denominator) minus y+ (seven as the numerator over 2 as the denominator).
Solve for y by simplify both sides with inverse operations, then isolate the variable and you get, y = 3x - 1.

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3 years ago

When writing a number in scientific notation, how many digits should be to the left of the decimal point?

Illusion [34]

B. You can only have one becase your answer has to be greater then one and less then 10. So it cant be 1 or 10 it has to be 2-9

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3 years ago

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How do you rationalize the numerator in this problem?

maw [93]

To solve this problem, you have to know these two special factorizations:

$x^3-y^3=(x-y)(x^2+xy+y^2)\\ x^3+y^3=(x+y)(x^2-xy+y^2)$

Knowing these tells us that if we want to rationalize the numerator. we want to use the top equation to our advantage. Let:

$\sqrt[3]{x+h}=x\\ \sqrt[3]{x}=y$

That tells us that we have:

$\frac{x-y}{h}$

So, since we have one part of the special factorization, we need to multiply the top and the bottom by the other part, so:

$\frac{x-y}{h}*\frac{x^2+xy+y^2}{x^2+xy+y^2}=\frac{x^3-y^3}{h*(x^2+xy+y^2)}$

So, we have:

$\frac{x+h-h}{h(\sqrt[3]{(x+h)^2}+\sqrt[3]{(x+h)(x)}+\sqrt[3]{x^2})}=\\ \frac{x}{\sqrt[3]{(x+h)^2}+\sqrt[3]{(x+h)(x)}+\sqrt[3]{x^2}}$

That is our rational expression with a rationalized numerator.

Also, you could just mutiply by:

$\frac{1}{\sqrt[3]{x_h}-\sqrt[3]{x}} \text{ to get}\\ \frac{1}{h\sqrt[3]{x+h}-h\sqrt[3]{h}}$

Either way, our expression is rationalized.

7 0

4 years ago

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Dmitriy789 [7]

409,380,000
Hope it helps

6 0

3 years ago

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