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3 years ago

8

Olivia has an average score 90 on her five French tests if she earns a score of 98 on the sixth test what will he average score

be

1 answer:

SVETLANKA909090 [29]3 years ago

7 0

Answer:

91.33

Step-by-step explanation:

90x5=450, add 98 to get 548. Since there are 6 tests, divide 548 by 6 and you get 91.33

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Convert the measurements 66,597 grams = how many kgs

Nuetrik [128]

Answer

66597 g

divide by 1000

66.597 kg

3 0

3 years ago

Quadrilateral OPQR is inscribed inside a circle as shown below. What is the measure of angle O? You must show all your work.

likoan [24]

Answer:

88 deg

Step-by-step explanation:

Opposite angles of an inscribed quadrilateral are supplementary. That means that the sum of the measures of angles O and Q is 180 deg.

2x + 2x + 4 = 180

4x + 4 = 180

4x = 176

x = 44

m<O = 2x = 2(44) = 88

Answer: 88 deg

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2 years ago

(40 POINTS) Line m is parallel to line n. The measure of angle 4 is (9a + 28)°. The measure of angle 6 is (10a)°. What is the me

Tems11 [23]

Answer:

$B. 100$

Step-by-step explanation:

This is simple, the picture that is provided tells us the answer easily.

An angle lower than 90 degrees is called an acute angle. While one that is the same as 90 degrees is a right angle, and one that is more than 90 is obtuse.

We are asked to take a look at angle 4, we can see that is more than 90 degrees. Since 100 is the only answer with an amount higher than 90 degrees and fits the angle, it is the angle.

8 0

3 years ago

Find the following integral

ololo11 [35]

There's nothing preventing us from computing one integral at a time:

$\displaystyle \int_0^{2-x} xyz \,\mathrm dz = \frac12xyz^2\bigg|_{z=0}^{z=2-x} \\\\ = \frac12xy(2-x)^2$

$\displaystyle \int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy = \frac12\int_0^{1-x}xy(2-x)^2\,\mathrm dy \\\\ = \frac14xy^2(2-x)^2\bigg|_{y=0}^{y=1-x} \\\\= \frac14x(1-x)^2(2-x)^2$

$\displaystyle\int_0^1\int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy\,\mathrm dx = \frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx$

Expand the integrand completely:

$x(1-x)^2(2-x)^2 = x^5-6x^4+13x^3-12x^2+4x$

Then

$\displaystyle\frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx = \left(\frac16x^6-\frac65x^5+\frac{13}4x^4-4x^3+2x^2\right)\bigg|_{x=0}^{x=1} \\\\ = \boxed{\frac{13}{240}}$

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2 years ago

The key idea of the transformation called a reflection is the rotating of a plane about a fixed point. true or false

Dennis_Churaev [7]

Answer: The answer is true your welcome

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers