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2 years ago

1. What is an extraneous solution of an equation?

1 answer:

6 0

“Extraneous solutions are not solutions at all. They arise from outside the problem, from the method of solution. They are extraneous because they are not solutions of the original problem. ... To tell if a "solution" is extraneous you need to go back to the original problem and check to see if it is actually a solution.”

Basically, they are an answer you get when you solve a problem, but they don’t actually make the equation true so you disregard it.

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Which sentence can be represented by the equation 3/5m=15

NeTakaya

Three- Fifth of m is equal to 15

Step-by-step explanation:

Step 1:

Here the expression given is

3/5 m = 15

Step 2:

To express the equation in terms of sentence,

THREE-FIFTH OF M IS EQUAL TO 15

6 0

2 years ago

What is the reference angle in radians for 1380°?

ad-work [718]

Hello there.

What is the reference angle in radians for 1380°?

Answer: 60

8 0

2 years ago

What is the answer of -32/40 + -35/40 ?

Alex787 [66]

-32/40 + -35/40 = -32+-35/40
= -67/40
= -1.675

The answer would then be -1.675.

3 0

3 years ago

A cube-shaped gift box has a volume of 5.12 cubic inches. What is the side length of the box?

kati45 [8]

recall that a cube has all equal sides, check the picture below.

$\bf \textit{volume of a cube}\\\\
V=x^3~~
\begin{cases}
x=side's~length\\[-0.5em]
\hrulefill\\
V=5.12
\end{cases}\implies 5.12=x^3
\\\\\\
\sqrt[3]{5.12}=x\implies 1.72354775\approx x$

6 0

2 years ago

Find cos(2*ABC) 100POINTS

juin [17]

Answer:

$-\dfrac{7}{25}$

Step-by-step explanation:

Trigonometric Identities

$\cos(A \pm B)=\cos A \cos B \mp \sin A \sin B$

Trigonometric ratios

$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$

where:

$\theta$ is the angle
O is the side opposite the angle
A is the side adjacent the angle
H is the hypotenuse (the side opposite the right angle)

Using the trig ratio formulas for cosine and sine:

$\cos(\angle ABC)=\dfrac{3}{5}$

$\sin(\angle ABC)=\dfrac{4}{5}$

Therefore, using the trig identities and ratios:

$\begin{aligned}\implies \cos(2 \cdot \angle ABC) & = \cos(\angle ABC + \angle ABC)\\\\& = \cos (\angle ABC) \cos (\angle ABC) - \sin(\angle ABC) \sin (\angle ABC)\\\\& = \cos^2(\angle ABC)-\sin^2(\angle ABC)\\\\& = \left(\dfrac{3}{5}\right)^2-\left(\dfrac{4}{5}\right)^2\\\\& = \dfrac{3^2}{5^2}-\dfrac{4^2}{5^2}\\\\& = \dfrac{9}{25}-\dfrac{16}{25}\\\\& = \dfrac{9-16}{25}\\\\& = -\dfrac{7}{25} \end{aligned}$

7 0

2 years ago

Read 2 more answers