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3 years ago

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the measure of one of the angles of a triangle is 35 degrees, the sum of the measures of the other two angles is 145 degrees and

the difference between their measures is 15 degrees. what are the measures of the unknown angles?

1 answer:

kozerog [31]3 years ago

7 0

80 and 65 are the measurements

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suzie spent 3 1/2 hours at the beach. she spent 45 minutes playing in the sand and the rest of the time playing in the water. ho

Anika [276]

Answer:

2 hours and 45 min

Explanation:

3 hours+30 minutes

subtract 30

3 hours

subtract 15

2 hours and 45 min

6 0

3 years ago

Help:) (no links please)!!!!

ICE Princess25 [194]

The answer is 9 miles!

3 0

3 years ago

Simplify: -|-15|<br><br> please answer correctly i'll give brainliest and 5-star rating.

12345 [234]

Answer:

-15

Step-by-step explanation:

<u>Definition of absolute value: </u>

<em>Distance from zero on a number line. </em>

<em />

Imagine a number line with 0 in the middle of it. Now mark the number 3 on it. Also mark the <em>opposite </em>of 3, which is -3. What is the <u>distance</u> from zero to those 2 points? it's 3.

This is exactly like saying |3| = 3 and |-3| = 3

Even if the number inside absolute value is negative, it <em><u>simplifies</u></em> to the positive of the same number.

_______________________________

So let's talk about your problem.

-|-15|

> the absolute value simplifies to 15

-|-15| = - 15

So that ends up being your answer.

$-15$

Hope this helps!!!!!!!

7 0

3 years 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

3 years ago

Find the greatest common factor of

labwork [276]

Answer:

D. xy.

Step-by-step explanation:

The GCF of x and x^2 is x and GCF of y and y^2 is y.

Answer is xy.

5 0

3 years ago

Read 2 more answers