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2 years ago

Help please!!!! thank you in advance. asap too

Mathematics

1 answer:

qaws [65]2 years ago

4 0

You can check the following
C=4
C=5
C=11

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Write the polynomial in standard form. Then name the polynomial based on its degree and number of terms.

sattari [20]

Let A= 2 – 11x2 – 8x + 6x2, the standard form of A is
A = - 5x² -8x + 2

7 0

2 years ago

Read 2 more answers

-8z+(4.5)+3.5z+7y-1.5

bogdanovich [222]

Answer:

$\boxed{\sf{7y+3-4.5z}}$

Step-by-step explanation:

In order to combine like terms, you have to isolate x and y from one side of the equation.

$\sf{-8z+\left(4.5\right)+3.5z+7y-1.5}$

First, thing you do is remove parentheses.

$\Longrightarrow: \sf{-8z+4.5+3.5z+7y-1.5}$

Solve.

Then, you combine like terms.

$\Longrightarrow:\sf{-8z+3.5z+7y+4.5-1.5}$

Add/subtract the numbers from left to right.

-8z+3.5z=-4.5z

Rewrite the problem down.

$\Longrightarrow: \sf{-4.5z+7y+4.5-1.5}$

Solve.

4.5-1.5=3

$\Longrightarrow: \boxed{\sf{7y+3-4.5z}}$

Therefore, the correct answer is 7y+3-4.5z.

I hope this helps, let me know if you have any questions.

6 0

1 year ago

Determine whether the following lines represented by the vector equations below intersect, are parallel, are skew, or are identi

KiRa [710]

Answer:

r(t) and s(t) are parallel.

Step-by-step explanation:

Given that :

the lines represented by the vector equations are:

r(t)=⟨1−t,3+2t,−3t⟩

s(t)=⟨2t,−3−4t,3+6t⟩

The objective is to determine if the following lines represented by the vector equations below intersect, are parallel, are skew, or are identical.

NOTE:

Two lines will be parallel if $\dfrac{x_1}{x_2}= \dfrac{y_1}{y_2}= \dfrac{z_1}{z_2}$

here;

$d_1 = (-1, \ 2, \ -3)$

Thus;

$r(t) = \dfrac{x-1}{-1} = \dfrac{y-3}{2}=\dfrac{z-0}{-3} = t$

$d_2 =(2, \ -4, \ +6)$

$s(t) = \dfrac{x-0}{2} = \dfrac{y+5}{-4}=\dfrac{z-3}{6} = t$

∴

$\dfrac{d_1}{d_2}= \dfrac{-1}{2} = \dfrac{2}{-4}= \dfrac{-3}{-6}$

Hence, we can conclude that r(t) and s(t) are parallel.

6 0

2 years ago

Please help me this is. big

kramer

Answer:

C. = 69.5

Step-by-step explanation:

The median or the average between 80 and 59 is 69.5

Hope this helps! :)

8 0

2 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

Help please!!!! thank you in advance. asap too​

Help please!!!! thank you in advance. asap too