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3 years ago

Question below :))))

Mathematics

2 answers:

svet-max [94.6K]3 years ago

7 0

1) x = 8

2) x = 50 degrees

Hope I helped! Thanks for posting your question! :)

8_murik_8 [283]3 years ago

3 0

The first answer is 8 the second is 50 degrees

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Use any method to add or subtract 5/7 {3/14+3/14}

OlgaM077 [116]

Answer:

15/49

Step-by-step explanation:

5/7 (3/14 + 3/14)

first combine like terms

3/14 + 3/14 = 6/14

then simplify by 2

6/14 to 3/7

now that both the term inside and outside have the same denominator, multiply

5/7 x 3/7 = 15/49

4 0

3 years ago

Determine the value of z in the figure answers: z = 50° z = 45° z = 10° z = 30°

ycow [4]

Answer:

z = 10°

Step-by-step explanation:

Since the two sides are equal, the two base angles have to be equal

That means that 5z and 130 form a straight line and add to 180 degrees

5z+130 = 180

Subtract 130 from each side

5z = 50

Divide by 5

5z/5 = 50/5

z=10

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3 years ago

Which conversion factors are used to convert 12 m/s to kilometers per minute?

Lera25 [3.4K]

Answer:

see the procedure

Step-by-step explanation:

we have

$12\ \frac{m}{s}$

Remember that

$1\ km=1,000\ m$

To convert m to km multiply by (1/1,000)

$1\ min=60\ sec$

To convert sec to min multiply by (1/60)

substitute

$12\ (\frac{(1/1,000)}{(1/60)})=12(\frac{60}{1,000})=0.72\ \frac{km}{min}$

7 0

3 years ago

What is 1,295.014 rounded to the nearest hundredth?

Sati [7]

1,295.01 is the correct answer

8 0

4 years ago

Read 2 more answers

Please solve i will give brainiest 100 point question ****** do the whole page please need to pass or i will fail its my final t

dmitriy555 [2]

Answer:

1. Find the difference between the areas.

Area of the small rectangle: $(x+2)(x+7)=x^2+7x+2x+14=x^2+9x+14$

Area of the big rectangle: $(x+9)(x+11)=x^2+11x+9x+99=x^2+20x+99$

The difference is: $11x+85$

$( x^2+20x+99)- (x^2+9x+14)=x^2+20x+99-x^2-9x-14=11x+85$

You can solve this question just by looking at the graph.

a) The height is 4 meters.

$f(d)=h=-2d^2+7d+4$

To find the height of the bleachers, we should consider the moment before the shoot, when the distance is equal to 0.

$f(0)=h=-2(0)^2+7(0)+4$

$h=4$

The height is 4 meters.

b) 9 meters.

For $d=1$

$f(1)=h=-2(1)^2+7(1)+4$

$f(1)=h=-2+7+4$

$h=9$

b) The ball travels 4 meters.

But to calculate it, it is when $h=0$

$0=-2d^2+7d+4$

Using the quadratic formula:

$d=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$d=\frac{-7 \pm \sqrt{7^2-4\left(-2\right)4}}{2\left(-2\right)}$

$d=\frac{-7\pm\sqrt{81}}{-4}$

$d=\frac{-7\pm9}{-4}$

It will give us to solutions, once it is a quadratic equation, but we are talking about a positive distance.

$d=-\frac{1}{2} \text{ or }d=4$

In this question, we have to find the area of the cylinder and the sphere.

From the information given, we have

a = 5mm and d = 5mm, therefore the radius is 2.5 mm.

The volume of a cylinder:

$V=\pi r^2h$

$V=\pi (2.5)^2 \cdot 5$

$V=31.25 \pi$

$V_{c} \approx 98.17 \text{ m}^3$

The volume of the sphere:

$V=\frac{4}{3} \pi r^2$

$V_{s} \approx 65.4 \text{ m}^3$

The volume of the capsule is approximately $163.57 \text{ m}^3$

3 0

4 years ago