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2 years ago

PLEASE help. I will mark as brainliest

Mathematics

2 answers:

finlep [7]2 years ago

4 0

Answer:

C 35

Step-by-step explanation:

Brainliest

blagie [28]2 years ago

3 0

The answer is 35! If you follow the pattern by moving over one and up one, then you’ll get to 35 after 7 hours.

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Solve completely (all missing sides and all missing angles) using trig ratios or laws. Round to the nearest tenth.

hodyreva [135]

Answer:

Step-by-step explanation:

3. Use Cosine law to find the length of the unknown side (PR)

q² = p² + r² - 2prCos Q

q is the opposite side of ∠Q;

p is the opposite side of ∠P; p = 33

r is the opposite sides of ∠R ; r = 67

q² = 33² + 67² - 2* 33*67 Cos 19°

= 1089 + 4489 - 4422 * 0.95

= 1089 + 4489 - 4200.9

= 1377.1

q = √1377.1

q = 37.1

PR = 37.1

To find the angle use law of sin

$\sf \dfrac{q}{Sin \ Q} =\dfrac{p}{Sin \ P}\\\\\\\dfrac{37.1}{Sin \ 19}=\dfrac{33}{Sin \ P} \\\\\\\dfrac{37.1}{0.33}=\dfrac{33}{Sin \ P}\\\\\\37.1* Sin \ P = 33 *0.33\\\\Sin \ P =\dfrac{33*0.33}{37.1}$

Sin P = 0.3

$P = Sin^{-1} 0.3\\$

P = 17.5°

∠R = 180 - (19 + 17.5)

= 143.5°

6 0

1 year ago

What is 1-cos(6x)=___

FromTheMoon [43]

Answer:

2sin^2(3x)

Step-by-step explanation:

3 0

1 year ago

Solve dis attachment and show all work ( I got it all wrong and I want to know how to solve it )

DedPeter [7]

(a) First find the intersections of $y=e^{2x-x^2}$ and $y=2$ :

$2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}$

So the area of $R$ is given by

$\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx$

If you're not familiar with the error function $\mathrm{erf}(x)$ , then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.

(b) Find the intersections of the line $y=1$ with $y=e^{2x-x^2}$ .

$1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2$

So the area of $S$ is given by

$\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx$
$\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx$

which is approximately 1.546.

(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve $y=e^{2x-x^2}$ and the line $y=1$ , or $e^{2x-x^2}-1$ . The area of any such circle is $\pi$ times the square of its radius. Since the curve intersects the axis of revolution at $x=0$ and $x=2$ , the volume would be given by

$\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx$

5 0

2 years ago

Evaluate x=-3, y=4, z=-5 Answers suggested below -17 41 23 -40 -58 -16

tatiyna

Answer:

2. -16

6. 23

7. -40

8. -17

9. 41

10. -58

6 0

2 years ago

Find the value of 50-(-15×-8

hoa [83]

Answer:

-70

Step-by-step explanation:

7 0

2 years ago