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3 years ago

How do you solve this algebreically? 3/8=x/20

Mathematics

1 answer:

Llana [10]3 years ago

6 0

Answer:

7.5

Step-by-step explanation:

3/8 = x/20

3(20) = 8x

8x = 60

x = 60/8

x = 7.5

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Determine if triangle ABC with coordinates A (0, 2), B (2, 5), and C (−1, 7) is an isosceles triangle. Use evidence to support y

kirill115 [55]

Answer:

The triangle ABC is an isosceles right triangle

Step-by-step explanation:

we have

The coordinates of triangle ABC are

A (0, 2), B (2, 5), and C (−1, 7)

we know that

An isosceles triangle has two equal sides and two equal internal angles

The formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

step 1

Find the distance AB

substitute in the formula

$d=\sqrt{(5-2)^{2}+(2-0)^{2}}$

$d=\sqrt{(3)^{2}+(2)^{2}}$

$dAB=\sqrt{13}\ units$

step 2

Find the distance BC

substitute in the formula

$d=\sqrt{(7-5)^{2}+(-1-2)^{2}}$

$d=\sqrt{(2)^{2}+(-3)^{2}}$

$dBC=\sqrt{13}\ units$

step 3

Find the distance AC

substitute in the formula

$d=\sqrt{(7-2)^{2}+(-1-0)^{2}}$

$d=\sqrt{(5)^{2}+(-1)^{2}}$

$dAC=\sqrt{26}\ units$

step 4

Compare the length sides

$dAB=\sqrt{13}\ units$

$dBC=\sqrt{13}\ units$

$dAC=\sqrt{26}\ units$

$dAB=dBC$

therefore

Is an isosceles triangle

Applying the Pythagoras Theorem

$(AC)^{2} =(AB)^{2}+(BC)^{2}$

substitute

$(\sqrt{26})^{2} =(\sqrt{13})^{2}+(\sqrt{13})^{2}$

$26=13+13$

$26=26$ -----> is true

therefore

Is an isosceles right triangle

8 0

3 years ago

A silo is not the shape of a cone. The silo is 8 meters tall and it's base has a diameter of 3 meters. Soybeans cost $414 per cu

Gnom [1K]

Answer:

$7803.72

Step-by-step explanation:

We have been given that a silo is in the shape of a cone. The silo is 8 meters tall and it's base has a diameter of 3 meters. Soybeans cost $414 per cubic meter. We are asked to find the total cost to fill the silo with soybeans.

First of all, we will find the volume of silo using volume of cone formula.

$\text{Volume of cone}=\frac{1}{3}\pi r^2 h$ , where,

r = radius

h = Height

We know that radius is half the diameter, so radius of silo would be $r=\frac{3}{2}=1.5$ .

$\text{Volume of cone}=\frac{1}{3}\pi (1.5\text{ m})^2 (8)\text{ m}$

$\text{Volume of cone}=\frac{1}{3}\pi\times 2.25\text{ m}^2 (8)\text{ m}$

$\text{Volume of cone}=\pi\times 0.75\text{ m}^2 (8)\text{ m}$

$\text{Volume of cone}=\pi\times 6\text{ m}^3$

$\text{Volume of cone}=18.8495559\text{ m}^3$

Now we will multiply total volume by $414 to find total cost.\

$\text{Total cost of soybeans}=\$414 \times 18.8495559$

$\text{Total cost of soybeans}=\$7803.7161426$

$\text{Total cost of soybeans}\approx \$7803.72$

Therefore, it will cost $7803.72 to fill the silo with soybeans.

5 0

3 years ago

Plsss hElpp!!!!! TY!!

kari74 [83]

Given:

A circle with diameter 9 cm.

To find:

The area of the circle.

Solution:

The area of a circle is:

$A=\pi r^2$ ...(i)

Where r is the radius.

We know that, the diameter of a circle is twice than its radius.

The diameter of the given circle is 9 cm.

$r=\dfrac{9}{2}$

$r=4.5$

Substituting $r=4.5, \pi =3.14$ in (i), we get

$A=(3.14)(4.5)^2$

$A=(3.14)(20.25)$

$A=63.585$

$A\approx 63.6$

Therefore, the area of the given circle is 63.6.

4 0

3 years ago

Can someone please help me with this?

geniusboy [140]

Answer: B and D

Step-by-step explanation:

First, solve the terms inside of the parenthesis. As a general rule, whenever you multiply two terms that have the same base, you can add their exponents.

Applying this rule, the base in the problem is 6, and the exponents are 3 and -4. The sum of 3 and -4 leaves -1. Therefore, this is one of our solutions:

$(6^-^1)^-^3$

When you have an exponential term raised to another exponent, you can simply multiply the exponents. In the previous solution given above, multiply -1 and -3 to get 3. Therefore our second solution is:

$6^3$

8 0

3 years ago

Find [5(cos 330 degrees + I sin 330 degrees)]^3

earnstyle [38]

Given a complex number in the form:
$z= \rho [\cos \theta + i \sin \theta]$
The nth-power of this number, $z^n$ , can be calculated as follows:

- the modulus of $z^n$ is equal to the nth-power of the modulus of z, while the angle of $z^n$ is equal to n multiplied the angle of z, so:
$z^n = \rho^n [\cos n\theta + i \sin n\theta ]$
In our case, n=3, so $z^3$ is equal to
$z^3 = \rho^3 [\cos 3 \theta + i \sin 3 \theta ] = (5^3) [\cos (3 \cdot 330^{\circ}) + i \sin (3 \cdot 330^{\circ}) ]$ (1)
And since
$3 \cdot 330^{\circ} = 990^{\circ} = 2\pi +270^{\circ}$
and both sine and cosine are periodic in $2 \pi$ , (1) becomes
$z^3 = 125 [\cos 270^{\circ} + i \sin 270^{\circ} ]$

6 0

3 years ago