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3 years ago

14

17 divided by 6294 idk what it is helllp

2 answers:

Oliga [24]3 years ago

7 0

Ummmmm is that the question or is there other numbers

Anastasy [175]3 years ago

6 0

The answer is 370.2352941

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Find the surface area of the regular pyramid

love history [14]

Greetings from Brazil!

$4\dfrac{1}{2} =\dfrac{9}{2}$

$Af1=\dfrac{b\cdot h}{2} \cdot 4\\\\\\Af1=\dfrac{\dfrac{9}{2} \cdot 4}{2} \cdot 4\\\\\\Af1=\dfrac{\dfrac{36}{2} }{2} \cdot 4\\\\\\Af1=\dfrac{18}{2} \cdot 4\\\\Af1=9\cdot 4\\\\Af1=36~m^2$

$Af2=l^2\\\\Af2=(\dfrac{9}{2})^2\\\\\ Af2=\dfrac{81}{4}\\\\Af2=20,25~m^2$

$At=Af1+Af2\\At=36+20,25\\At=56,25~m^2$

I hope I helped you! =)

6 0

3 years ago

Draw an angle XYZ of 88 degree . Construct the angle bisector of

sergejj [24]

Answer:

Step-by-step explanation:

An angle is formed where two or more lines meet. It is measured in degrees. And an angle bisector is a straight line that divides an angle into two equal parts.

Given: <XYZ = $88^{0}$ .

Construction: A bisector of <XYZ.

The construction can be seen in the attached diagram.

8 0

3 years ago

The pet doctor weighs Marc's dog every year. Last year, the dog lost 3.73 pounds from the previous year. This year, the dog gain

levacccp [35]

Answer:

-2.26 lb

Step-by-step explanation:

Let the dog's initial weight be d.

Then, one year later, the dog weighs 3.73 lb less, and thus weighed d - 3.73 lb.

During the next year, the dog gained 1.47 lb, so now weighs d - 3.73 + 1.47 lb, or

d - 2.26 lb. This indicates a total weight loss of 2.26 over these two years.

6 0

3 years ago

The ball's height( in meters above the ground), x seconds after Amir threw it, is modeled by : h(x)=-(x-2)^2 +16 what is the hei

Bess [88]

Answer:

20 m

Step-by-step explanation:

The height of the ball in metres is given as:

$h(x) = (x-2)^2 + 16$

where xis time, in seconds.

At the instant that the ball is thrown, the time is 0 seconds. This means that the height will be h(0):

$h(0) = (0 - 2)^2 + 16\\\\\\h(0) = (-2)^2 + 16\\\\\\h(0) = 4 + 16\\\\\\h(0) = 20 m$

The height of the ball is 20 metres.

3 0

4 years ago

Read 2 more answers

Given a data set has a Median of 10 and an Inner Quartile Range of 5, what is the range of values that Q3 could possibly be?

Elanso [62]

Because the Median could be anywhere within the interquartile range the upper quartile could be between 10 and 15

5 0

3 years ago

Read 2 more answers