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wariber [46]
3 years ago
8

Need help on math again plz

Mathematics
2 answers:
EastWind [94]3 years ago
7 0

Answer:

Socratic app

Step-by-step explanation:

it will help you

aliya0001 [1]3 years ago
7 0

Answer:

I can't see the image

Step-by-step explanation:

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What is the area of this figure?
Lorico [155]
Each little square is 1 inch high and 1 inch wide, so
the area of each little square is 1 square inch.

The bug figure is made of 23 little squares, so the area
of the big figure is 23 square inches.
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3 years ago
Part A: Explain why the x-coordinates of the points where the graphs of the equations y = 4-x and y = 2x + 3 intersect are the s
scoray [572]
Hope it cleared your doubt.

7 0
4 years ago
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The mass is divide by 2 and volume multiples by 4. What happens to the density?
leva [86]

Answer:

Step-by-step explanation:

Density is an intensive property of the material or substance and depends upon the relationship between the mass and volume. Unless the mass changes in relation to the volume, the density will not change.

5 0
3 years ago
Please help me for the love of God if i fail I have to repeat the class
Elena-2011 [213]

\theta is in quadrant I, so \cos\theta>0.

x is in quadrant II, so \sin x>0.

Recall that for any angle \alpha,

\sin^2\alpha+\cos^2\alpha=1

Then with the conditions determined above, we get

\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35

and

\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}

Now recall the compound angle formulas:

\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta

\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta

\sin2\alpha=2\sin\alpha\cos\alpha

\cos2\alpha=\cos^2\alpha-\sin^2\alpha

as well as the definition of tangent:

\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}

Then

1. \sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}

2. \cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}

3. \tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}

4. \sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}

5. \cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}

6. \tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7

7. A bit more work required here. Recall the half-angle identities:

\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2

\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2

\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}

Because x is in quadrant II, we know that \dfrac x2 is in quadrant I. Specifically, we know \dfrac\pi2, so \dfrac\pi4. In this quadrant, we have \tan\dfrac x2>0, so

\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32

8. \sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}

6 0
4 years ago
How many grams of active ingredient is contained in 500ml of a 30% solution
netineya [11]

let's reword it, since the wording is crummy

"How many grams of active ingredient is contained in 500mL of a solution which is 30% of the active ingredient?"

namely, what is 30% of 500?

\begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{30\% of 500}}{\left( \cfrac{30}{100} \right)500}\implies 150~mL

4 0
2 years ago
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