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Marta_Voda [28]

3 years ago

11

PLEASE HELP ASAPPPPP THANK YOU

1 answer:

Arada [10]3 years ago

8 0

Answer:

i wish i could help u!!!!!!

Step-by-step explanation:

uuuhhhhh i hate that i kinda suck at math

anyway, i'll give this my best shot

The midsegment of a triangle is a line segment that connects the midpoints of two sides ofrigidity to structures. In the support for the image shown shown, the crossbar GJK is a midsegment of GFK

You can use a compass and straightedge to construct the midsegments of a triangle.

Hope this helps! It's the best i can do!

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a sailfish can go as fast as 68mph. in 1 min. can a sailfish swim as far as 1 mile? explain the answer.

Bezzdna [24]

Yes the sailfish can swim one mile

7 0

4 years ago

Mia wants to buy 3 notebooks for $1.29 each. What is the expression for the total cost?

ololo11 [35]

Step-by-step explanation:

Mia want to buy 3 notebook for $1.29 each ,

then total cost =3×1.29

hence, 3×29

1\6+1/12+2/6=2+1+4/12

=7/12

8 0

3 years ago

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.

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

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

The difference is: $11x+85$

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

2.

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.

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

3.

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

Solve for x<br>-8+6(-x-3)+4x

zalisa [80]

-x-26 This is the answer

6 0

3 years ago

alexandr402 [8]

Answer:

let the grade points be a,b,c

a+b+c=258

b=c+10( since she noticed that on tests two and three,she scored 10 points fewer).

3 0

3 years ago

Read 2 more answers