Mark filed a claim after his car was rear-ended by another driver. Which

One way of solving systems of linear equations is by adding a multiple of one equation to the other. the multiplier for one equa

elena-s [515]

We want to use elimination to solve
y = 0 (1)
x + y = 40 (2)

Multiply (1) by -1 to eliminate y.
-y = 0 (3)

Add (2) and (3).
x + y + (-y) = 40 + 0
x = 40

Answer: Multiply by -1.

3 0

2 years ago

Let f(x) = x - 3 and g(x) = x + 11. Find f(x) · g(x).

Nady [450]

Answer:

i think its D

f(x) = x-3 and g(x) = x+11

f(x) *g(x) = (x-3)(x+11) = x^2 +11x -3x -33 = x^2 +8x -33

hope this helped :3

3 0

2 years ago

Read 2 more answers

tapes-r-us is have a sale on hq tape players. The regular price is $60. The discount is $18. What percent discount is this?

olga nikolaevna [1]

The discount is 30%

Step-by-step explanation:

Given

Price of tape = $60

Discount = $18

In order to find the percentage, we will use the percentage formula

$Percentage = \frac{discount}{Price}*100\\= \frac{18}{60} *100\\=0.3 * 100\\= 30\%$

Hence,

The discount is 30%

Keywords: Percent, sale

Learn more about percentage at:

#LearnwithBrainly

7 0

2 years ago

Find the remaining trigonometric ratios of θ if csc(θ) = -6 and cos(θ) is positive

VikaD [51]

Now, the cosecant of θ is -6, or namely -6/1.

however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.

we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

$\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}$

recall that

$\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad\qquad 
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\\\\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\\\\\\
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\qquad \qquad 
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}$

therefore, let's just plug that on the remaining ones,

$\bf sin(\theta)=\cfrac{-1}{6}
\qquad\qquad 
cos(\theta)=\cfrac{\sqrt{35}}{6}
\\\\\\
% tangent
tan(\theta)=\cfrac{-1}{\sqrt{35}}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{\sqrt{35}}{1}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}$

now, let's rationalize the denominator on tangent and secant,

$\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}$

3 0

2 years ago

The decimal equivalent of is 3/4 is...

tatuchka [14]

Answer:

0.75

Step-by-step explanation:

since it's a fraction you can divide

3 divided by 4 is 0.75

4 0

1 year ago