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3 years ago

I NEED THIS ASAP!!! What is GC?

Mathematics

1 answer:

umka21 [38]3 years ago

6 0

9514 1404 393

Answer:

D. 14

Step-by-step explanation:

Point G divides each median into parts that have the ratio ...

short : long = 1 : 2

Then ...

GD : GC = 1 : 2 = 7 : 14 . . . . . . . . multiply the ratio by 7

GC = 14

You might be interested in

"Which word is used to name the bold numbers and variables?

Kay [80]

If the bold numbers and variables are 2x + 5, then those would be B. addend, or the numbers that are added. Minuend is the number from which you subtract.
And since there is no dividing here, C or D are incorrect.

7 0

3 years ago

Paul has three cube-shaped boxes. Each box is a different size

qwelly [4]

Answer:

See explanation

Step-by-step explanation:

Paul has three cube-shaped boxes. Each box is a different size and they are stacked from the largest to the smallest. Some information about the boxes is given below.

The combined volume of the three boxes is 1,197 cubic inches.
The area of one face of the medium box is 49 square inches.
The volume of the smallest box is 218 cubic inches less than the volume of the medium box.

1. The medium box has the area of one face of 49 square inches, then

$a^2=49\\ \\a=7\ inches$

is the side length.

The volume of the medium box is

$a^3=7^3=343\ in^3.$

2. The volume of the smallest box is 218 cubic inches less than the volume of the medium box, then the volume of the smallest box is

$343-218=125\ in^3.$

Ib is the side length, then

$b^3=125\\ \\b=5\ inches$

The area of one face is

$b^2=5^2=25\ in^2.$

3. The volume of the largest box is

$1,197-343-125=729\ in^3,$

then if c is the side length,

$c^3=729\\ \\c=9\ inches.$

4. The total height of the stack is the sum of all sides lengths:

$a+b+c=7+5+9=21\ inches$

5. Find the surface area of each box:

small $6b^2=6\cdot 5^2=6\cdot 25=150\ in^2;$
medium $6a^2=6\cdot 7^2=6\cdot 49=294\ in^2;$
large $6c^2=6\cdot 9^2=6\cdot 81=486\ in^2.$

In total, Paul needs

$150+294+486=930\ in^2$

of wrapping paper. He has 1,000 square inches, so Paul has enough paper to wrap all 3 boxes.

8 0

4 years ago

A piece of wire 19 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria

mr Goodwill [35]

Answer: 8.26 m

Step-by-step explanation:

$Let s be the length of the wire used for the square. \\Let $t$ be the length of the wire used for the triangle. \\Let $A_{S}$ be the area of the square. \\Let ${A}_{T}}$ be the area of the triangle. \\One side of the square is $\frac{s}{4}$ \\Therefore,we know that,$$A_{S}=\left(\frac{s}{4}\right)^{2}=\frac{s^{2}}{16}$

$The formula for the area of an equilateral triangle is, $A=\frac{\sqrt{3}}{4} a^{2}$ where $a$ is the length of one side,And one side of our triangle is $\frac{t}{3}$So,We know that,$$A_{T}=\frac{\sqrt{3}}{4}\left(\frac{t}{3}\right)^{2}$$We have to find the value of "s" such that,$\mathrm{s}+\mathrm{t}=19$ hence, $\mathrm{t}=19-\mathrm{s}$And$$A_{S}+A_{T}=A_{S+T}$

$$$Therefore,$$\begin{aligned}&A_{T}=\frac{\sqrt{3}}{4}\left(\frac{(19-s)}{3}\right)^{2}=\frac{\sqrt{3}(19-s)^{2}}{36} \\&A_{T+S}=\frac{s^{2}}{16}+\frac{\sqrt{3}(19-s)^{2}}{36}\end{aligned}$

$Differentiating the above equation with respect to s we get,$$A^{\prime}{ }_{T+S}=\frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}$$Now we solve $A_{S+T}^{\prime}=0$$$\begin{aligned}&\Rightarrow \frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}=0 \\&\Rightarrow \frac{s}{8}=\frac{\sqrt{3}(19-s)}{18}\end{aligned}$$Cross multiply,$$\begin{aligned}&18 s=8 \sqrt{3}(19-s) \\&18 s=152 \sqrt{3}-8 \sqrt{3} s \\&(18+8 \sqrt{3}) s=152 \sqrt{3} \\&s=\frac{152 \sqrt{3}}{(18+8 \sqrt{3})} \approx 8.26\end{aligned}$

$The domain of $s$ is $[0,19]$.So the endpoints are 0 and 19$$\begin{aligned}&A_{T+S}(0)=\frac{0^{2}}{16}+\frac{\sqrt{3}(19-0)^{2}}{36} \approx 17.36 \\&A_{T+S}(8.26)=\frac{8.26^{2}}{16}+\frac{\sqrt{3}(19-8.26)^{2}}{36} \approx 9.81 \\&A_{T+S}(19)=\frac{19^{2}}{16}+\frac{\sqrt{3}(19-19)^{2}}{36}=22.56\end{aligned}$

$$$Therefore, for the minimum area, $8.26 \mathrm{~m}$ should be used for the square$

8 0

2 years ago

Find the volume of the cone round your answer to the nearest tenth

Ainat [17]

Answer:

209.43951 round to the nearest tenth is 209.44

Step-by-step explanation:

A cone with a base radius of 5 units and a height of 8 units has a volume of 209.44 cubed units.

If a cone has a flat bottom, meaning the height and radius meet at right angles, then this formula can be used to find of volume ('V') of that cone (also know as a right circular cone):

V = 1/3(PI*r squared * h )

The volume of a cone can be calculated by taking one-third of the result of the radius squared, multiplied by the height, multiplied by the mathematical constant pi.

Here is a step-by-step case that illustrates how to find to volume of a cone with a radius of 2 feet and a height of 3 feet. Note that in order to save space (and because pi cannot be determine precisely) a limited number of decimal places are used, the symbol ~ denotes that this answer is an approximation.

V = 1/3(pi * r^2 * h)

= 1/3(pi * 2^2 * 3)

= 1/3(pi * 12)

= 1/3(37.7)

~ 12.6 cubic feet

Hope this helps!!

6 0

3 years ago

Complete the problems. Show your work.

timama [110]

1. =4.8
2. =40%
3. = 160

8 0

3 years ago