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3 years ago

Persephone is creating a flower garden in her back yard. If she needs 30 pounds

Mathematics

1 answer:

Likurg_2 [28]3 years ago

5 0

I thinks it’s B but I’m not sure sorry if not but B

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Evaluate the iterated integral. $$ \int\limits_0^{2\pi}\int\limits_0^y\int\limits_0^x {\color{red}9} \cos(x+y+z)\,dz\,dx\,dy $$

KengaRu [80]

$\displaystyle\int_{y=0}^{y=2\pi}\int_{x=0}^{x=y}\int_{z=0}^{z=x}\cos(x+y+z)\,\mathrm dz\,\mathrm dx\,\mathrm dy=\int_{y=0}^{y=2\pi}\int_{x=0}^{x=y}\sin(x+y+z)\bigg|_{z=0}^{z=x}\,\mathrm dx\,\mathrm dy$
$\displaystyle=\int_{y=0}^{y=2\pi}\int_{x=0}^{x=y}\sin(2x+y)-\sin(x+y)\,\mathrm dx\,\mathrm dy$
$\displaystyle=\int_{y=0}^{y=2\pi}-\frac12\left(\cos(2x+y)-2\cos(x+y)\right)\bigg|_{x=0}^{x=y}\,\mathrm dx\,\mathrm dy$
$\displaystyle=\int_{y=0}^{y=2\pi}-\frac12\left((\cos3y-2\cos2y)-(\cos y-2\cos y)\right)\bigg|_{x=0}^{x=y}\,\mathrm dy$
$\displaystyle=-\frac12\int_{y=0}^{y=2\pi}(\cos3y-2\cos2y+\cos y)\,\mathrm dy$
$\displaystyle=-\frac12\left(\frac13\sin3y-\sin2y+\sin y\right)\bigg|_{y=0}^{y=2\pi}$
$=0$

4 0

3 years ago

Your math teacher tells you that the next test is worth 100 points and contains 38 problems. Multiple-choice questions are worth

erastova [34]

Answer:

Most likely 8 word problems and 30 multiple choice

Step-by-step explanation:

8x5=40 and 30×2=60

40+60=100

7 0

3 years ago

Analia is a school district manager. Here are some details about two schools in her district for the last school year:

AfilCa [17]

Answer is in this link:

brainly.com/question/17270742

6 0

2 years ago

PLS ANSWER ASAP 30 POINTS!!! CHECK PHOTO! WILL MARK BRAINLIEST TO WHO ANSWERS

Sveta_85 [38]

I'll do Problem 8 to get you started

a = 4 and c = 7 are the two given sides

Use these values in the pythagorean theorem to find side b

$a^2 + b^2 = c^2\\\\4^2 + b^2 = 7^2\\\\16 + b^2 = 49\\\\b^2 = 49 - 16\\\\b^2 = 33\\\\b = \sqrt{33}\\\\$

With respect to reference angle A, we have:

opposite side = a = 4
adjacent side = b = $\sqrt{33}$
hypotenuse = c = 7

Now let's compute the 6 trig ratios for the angle A.

We'll start with the sine ratio which is opposite over hypotenuse.

$\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(A) = \frac{a}{c}\\\\\sin(A) = \frac{4}{7}\\\\$

Then cosine which is adjacent over hypotenuse

$\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}\\\\\cos(A) = \frac{b}{c}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\$

Tangent is the ratio of opposite over adjacent

$\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(A) = \frac{a}{b}\\\\\tan(A) = \frac{4}{\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{\sqrt{33}*\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{(\sqrt{33})^2}\\\\\tan(A) = \frac{4\sqrt{33}}{33}\\\\$

Rationalizing the denominator may be optional, so I would ask your teacher for clarification.

So far we've taken care of 3 trig functions. The remaining 3 are reciprocals of the ones mentioned so far.

cosecant, abbreviated as csc, is the reciprocal of sine
secant, abbreviated as sec, is the reciprocal of cosine
cotangent, abbreviated as cot, is the reciprocal of tangent

So we'll flip the fraction of each like so:

$\csc(\text{angle}) = \frac{\text{hypotenuse}}{\text{opposite}} \ \text{ ... reciprocal of sine}\\\\\csc(A) = \frac{c}{a}\\\\\csc(A) = \frac{7}{4}\\\\\sec(\text{angle}) = \frac{\text{hypotenuse}}{\text{adjacent}} \ \text{ ... reciprocal of cosine}\\\\\sec(A) = \frac{c}{b}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(\text{angle}) = \frac{\text{adjacent}}{\text{opposite}} \ \text{ ... reciprocal of tangent}\\\\\cot(A) = \frac{b}{a}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\$

------------------------------------------------------

Summary:

The missing side is $b = \sqrt{33}$

The 6 trig functions have these results

$\sin(A) = \frac{4}{7}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\\tan(A) = \frac{4}{\sqrt{33}} = \frac{4\sqrt{33}}{33}\\\\\csc(A) = \frac{7}{4}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\$

Rationalizing the denominator may be optional, but I would ask your teacher to be sure.

7 0

1 year ago

3d - 4, where d = 5 .

Crank

The correct answer would be-11

8 0

3 years ago

Read 2 more answers