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2 years ago

9

Helpppp how do I do this?

1 answer:

zhannawk [14.2K]2 years ago

5 0

The angles shown are

<QPR
<SPR

PR is the bisector of angle P

So

<P=<QPR+<SPR or =<QPS

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Four different sets of objects contain 2,5,6, and 7 objects, respectively. How many unique combinations can be formed by picking

Ilia_Sergeevich [38]

There are 2 choices for the first set, and 5 choices for the second set. Each of the 2 choices from the first set can be combined with each of the 5 choices from the second set. Therefore there are 2 times 5 combinations from the first and second sets. Continuing this reasoning, the total number of unique combinations of one object from each set is:
$2\times5\times\times6\times7=420\ combinations$

8 0

3 years ago

A regression analysis between sales (y in $1000) and advertising (x in dollars) resulted in the following equation y = 50,000 +

Vikentia [17]

Answer:

D. increase of $1 in advertising is associated with an increase of $6,000 in sales.

Step-by-step explanation:

Every dollar in advertisting is multiply by 6 in the equation, but every unit in y represent $1000. That means that $1 1n advertisting is multiply by 6 and then by $1000, that is $1 in advertisting = $6000 in sales.

4 0

3 years ago

Please please please help and solve this with steps, much help needed thank you :) 20 points for this!

leonid [27]

Answer:

$y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}$

(Please vote me Brainliest if this helped!)

Step-by-step explanation:

$\frac{dy}{dx}y=x^3+2x-5x+3$

$\mathrm{First\:order\:separable\:Ordinary\:Differential\:Equation}$

$\mathrm{A\:first\:order\:separable\:ODE\:has\:the\:form\:of}\:N\left(y\right)\cdot y'=M\left(x\right)$

1. $\mathrm{Substitute\quad }\frac{dy}{dx}\mathrm{\:with\:}y'\:$

$y'\:y=x^3+2x-5x+3$

2. $\mathrm{Rewrite\:in\:the\:form\:of\:a\:first\:order\:separable\:ODE}$

$yy'\:=x^3-3x+3$

$N\left(y\right)\cdot y'\:=M\left(x\right)$

$N\left(y\right)=y,\:\quad M\left(x\right)=x^3-3x+3$

3. $\mathrm{Solve\:}\:yy'\:=x^3-3x+3:\quad \frac{y^2}{2}=\frac{x^4}{4}-\frac{3x^2}{2}+3x+c_1$

4. $\mathrm{Isolate}\:y:\quad y=\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}}$

5. $\mathrm{Simplify}$

$y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}$

7 0

3 years ago

Read 2 more answers

Tim has $250 in his account. This is $75 more than his brother Nate has in his account. Write and solve an addition equation to

BaLLatris [955]

Answer: 250+75

Step-by-step explanation:

250+75= 325

7 0

3 years ago

Read 2 more answers

Will reward brainliest!

Lina20 [59]

Option D: Two irrational solutions

Explanation:

The equation is $17+3 x^{2}=6 x$

Subtracting 6x from both sides, we have,

$3x^{2} -6x+17=0$

Solving the equation using quadratic formula,

$x=\frac{6 \pm \sqrt{36-4(3)(17)}}{2(3)}$

Simplifying the expression, we get,

$\begin{aligned}x &=\frac{6 \pm \sqrt{36-204}}{6} \\&=\frac{6 \pm \sqrt{-168}}{6} \\&=\frac{6 \pm 2 i \sqrt{42}}{6}\end{aligned}$

Taking out the common terms and simplifying, we have,

$\begin{aligned}x &=\frac{2(3 \pm i \sqrt{42})}{6} \\&=\frac{(3 \pm i \sqrt{42})}{3}\end{aligned}$

Dividing by 3, we get,

$x=1+i \sqrt{\frac{14}{3}}, x=1-i \sqrt{\frac{14}{3}}$

Hence, the equation has two irrational solutions.

8 0

3 years ago