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3 years ago

Which table represents a function

Mathematics

1 answer:

solong [7]3 years ago

4 0

Answer:

The last one

Step-by-step explanation:

A function is where every x has only one y value.

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Math Help

-Dominant- [34]

X=1, because they both have the same height (y), so they'll cross each other!

4 0

3 years ago

For what value of a should you solve the system of elimination?

SIZIF [17.4K]

$\begin{bmatrix}3x+5y=10\\ 2x+ay=4\end{bmatrix}$

$\mathrm{Multiply\:}3x+5y=10\mathrm{\:by\:}2: 6x+10y=20$
$\mathrm{Multiply\:}2x+ay=4\mathrm{\:by\:}3: 3ay+6x=12$

$\begin{bmatrix}6x+10y=20\\ 6x+3ay=12\end{bmatrix}$

6x + 3ay = 12
-
6x + 10y = 20
/
3a - 10y = -8

$\begin{bmatrix}6x+10y=20\\ 3a-10y=-8\end{bmatrix}$

$3a-10y=-8 \ \textgreater \ \mathrm{Subtract\:}3a\mathrm{\:from\:both\:sides}$
$3a-10y-3a=-8-3a$

$\mathrm{Simplify} \ \textgreater \ -10y=-8-3a \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}-10$
$\frac{-10y}{-10}=-\frac{8}{-10}-\frac{3a}{-10}$

Simplify more.

$\frac{-10y}{-10} \ \textgreater \ \mathrm{Apply\:the\:fraction\:rule}: \frac{-a}{-b}=\frac{a}{b} \ \textgreater \ \frac{10y}{10}$

$\mathrm{Divide\:the\:numbers:}\:\frac{10}{10}=1 \ \textgreater \ y$

$-\frac{8}{-10}-\frac{3a}{-10} \ \textgreater \ \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{-8-3a}{-10}$

$\mathrm{Apply\:the\:fraction\:rule}: \frac{a}{-b}=-\frac{a}{b} \ \textgreater \ -\frac{-3a-8}{10} \ \textgreater \ y=-\frac{-8-3a}{10}$

$\mathrm{For\:}6x+10y=20\mathrm{\:plug\:in\:}\ \:y=\frac{8}{10-3a} \ \textgreater \ 6x+10\cdot \frac{8}{10-3a}=20$

$10\cdot \frac{8}{10-3a} \ \textgreater \ \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \ \frac{8\cdot \:10}{10-3a}$
$\mathrm{Multiply\:the\:numbers:}\:8\cdot \:10=80 \ \textgreater \ \frac{80}{10-3a}$

$6x+\frac{80}{10-3a}=20 \ \textgreater \ \mathrm{Subtract\:}\frac{80}{10-3a}\mathrm{\:from\:both\:sides}$
$6x+\frac{80}{10-3a}-\frac{80}{10-3a}=20-\frac{80}{10-3a}$

$\mathrm{Simplify} \ \textgreater \ 6x=20-\frac{80}{10-3a} \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}6 \ \textgreater \ \frac{6x}{6}=\frac{20}{6}-\frac{\frac{80}{10-3a}}{6}$

$\frac{6x}{6} \ \textgreater \ \mathrm{Divide\:the\:numbers:}\:\frac{6}{6}=1 \ \textgreater \ x$

$\frac{20}{6}-\frac{\frac{80}{10-3a}}{6} \ \textgreater \ \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{20-\frac{80}{-3a+10}}{6}$

$20-\frac{80}{10-3a} \ \textgreater \ \mathrm{Convert\:element\:to\:fraction}: \:20=\frac{20}{1} \ \textgreater \ \frac{20}{1}-\frac{80}{-3a+10}$

$\mathrm{Find\:the\:least\:common\:denominator\:}1\cdot \left(-3a+10\right)=-3a+10$

$Adjust\:Fractions\:based\:on\:the\:LCD \ \textgreater \ \frac{20\left(-3a+10\right)}{-3a+10}-\frac{80}{-3a+10}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}: \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$
$\frac{20\left(-3a+10\right)-80}{-3a+10} \ \textgreater \ \frac{\frac{20\left(-3a+10\right)-80}{-3a+10}}{6} \ \textgreater \ \mathrm{Apply\:the\:fraction\:rule}: \frac{\frac{b}{c}}{a}=\frac{b}{c\:\cdot \:a}$

$20\left(-3a+10\right)-80 \ \textgreater \ Rewrite \ \textgreater \ 20+10-3a-4\cdot \:20$

$\mathrm{Factor\:out\:common\:term\:}20 \ \textgreater \ 20\left(-3a+10-4\right) \ \textgreater \ Factor\;more$

$10-3a-4 \ \textgreater \ \mathrm{Subtract\:the\:numbers:}\:10-4=6 \ \textgreater \ -3a+6 \ \textgreater \ Rewrite$
$-3a+2\cdot \:3$

$\mathrm{Factor\:out\:common\:term\:}3 \ \textgreater \ 3\left(-a+2\right) \ \textgreater \ 3\cdot \:20\left(-a+2\right) \ \textgreater \ Refine$
$60\left(-a+2\right)$

$\frac{60\left(-a+2\right)}{6\left(-3a+10\right)} \ \textgreater \ \mathrm{Divide\:the\:numbers:}\:\frac{60}{6}=10 \ \textgreater \ \frac{10\left(-a+2\right)}{\left(-3a+10\right)}$

$\mathrm{Remove\:parentheses}: \left(-a\right)=-a \ \textgreater \ \frac{10\left(-a+2\right)}{-3a+10}$

$Therefore\;our\;solutions\;are\; y=\frac{8}{10-3a},\:x=\frac{10\left(-a+2\right)}{-3a+10}$

Hope this helps!

7 0

3 years ago

Read 2 more answers

What is the slope of the line on the graph ?

Minchanka [31]

Answer is -2

Step-by step solution:

5 0

2 years ago

Pentagon RSTUV is circumscribed about a circle. What is the value of x if RS = 8, ST = 12, TU = 15, UV = 9, and VR = 12?

ivanzaharov [21]

Answer:

The value of x is 7 ⇒ 1st answer

Step-by-step explanation:

* Lets revise a fact in the circle

- The two tangents drawn from a point out side the circle are equal

∵ RSTUV is circumscribed about a circle

∴ Each side of the pentagon is a tangent to the circle

- Look to the attached figure to know how we will solve the problem

- Each tangent divided into two parts

# RS = x + y

∵ RS = 8

∴ x + y = 8 ⇒ (1)

# RV = x + n

∵ RV = 12

∴ x + n = 12 ⇒ (2)

- Subtract (2) from (1)

∴ y - n = -4 ⇒ (3)

# ST = y + z

∵ ST = 12

∴ y + z = 12 ⇒ (4)

# TU = z + m

∵ TU = 15

∴ z + m = 15 ⇒ (5)

- Subtract (5) from (4)

∴ y - m = -3 ⇒ (6)

# UV = m + n

∵ UV = 9

∴ m + n = 9 ⇒ (7)

- Add (6) and (7)

∴ y + n = 6 ⇒ (8)

- Lets solve equation (3) and equation (8) to find y

∵ y - n = -4 ⇒ (3)

∵ y + n = 6 ⇒ (8)

- Add (3) and (8)

∴ 2y = 2 ⇒ divide two sises by 2

∴ y = 1

- Lets substitute the value of y in equation (1)

∵ x + y = 8 ⇒ (1)

∵ y = 1

∴ x + 1 = 8 ⇒ subtract (1) from both sides

∴ x = 7

* The value of x is 7

3 0

3 years ago

Use the Factor Theorem to show that (x - 3) is a factor of the polynomial

Katyanochek1 [597]

Answer:

Step-by-step explanation:

<u>Factor Theorem is a consequence of Remainder Theorem. </u>

Remainder Theorem states that if polynomial f(x) is divided by a binomial (x - a) then the remainder is f(a).

Factor Theorem states that if f(a) = 0, then the binomial (x - a) is a factor of f(x).

We have the polynomial

$P(x) = x^5-3x^4+5x^3-15x^2-6x+18$

To prove that x-3 is a factor of P, we calculate P(3):

$P(3) = 3^5-3*3^4+5*3^3-15*3^2-6*3+18$

$P(3) = 243-243+135-135-18+18$

$P(3) = 0$

Thus, x-3 is a factor of P(x)

7 0

3 years ago