Which of the following exponential equations could be represented by the table below?

Will someone please help me!!! I don’t know the answer to these! I don’t even know if I did them right!

Varvara68 [4.7K]

Answer:

$BD = 12$ units

$AB = 16$ units

$FG=\sqrt{105}$

$EC=4\sqrt{10}$ units

$AC=8\sqrt{10}$ units

Step-by-step explanation:

A circumcenter is the point where the perpendicular bisectors of a triangle intersect.

This means

BD = DC;
BF = FA;
AE = EC.

Since BC = 24 units, then BD = 12 units (half of BC).

Since FA = 8 units, AB = 16 units (twice FA).

Consider right triangle BGF. In this triangle, BF = 8 units (congruent to FA), BG = 13 units. By the Pythagorean theorem,

$FG^2=BG^2-BF^2\\ \\FG^2=13^2-8^2\\ \\FG^2=169-64=105\\ \\FG=\sqrt{105}\ units$

Triangles BFG and AFG are congruent right triangles, then BG = AG. In right triangle AGE, $GE=3$ units, $AG=13$ units. By the Pythagorean theorem,

$AE^2=AG^2-GE^2\\ \\AE^2=13^2-3^2=169-9=160\\ \\AE=\sqrt{160}=4\sqrt{10}\ units$

$AE = EC=4\sqrt{10}\ units$

$AC = 2AE=2\cdot 4\sqrt{10}=8\sqrt{10}\ units$

5 0

3 years ago

Jodi built a rectangular sandbox for her daughter. The sandboxmeasures6 feet by 5feet by 1.2 feet. Jodi purchases 20cubic feet o

Mamont248 [21]

For this case, the first thing to do is to model the rectangular sandbox as a rectangular prism.

The volume of the prism is given by:

$V = w * l * h$

Where,

<em>w: width </em>
<em>l: length </em>
<em>h: height </em>

Therefore, replacing values we have:

$V = 6 * 5 * 1.2\\V = 36 ft ^ 3$

We observed that:

$36 ft ^ 3> 20 ft ^ 3$

Therefore, the sandbox is not completely filled.

Answer:

the sandbox will hold $36 ft ^ 3$ of sand

Jodi purchase is not enough sand to fill the sandbox to the top.

$36 ft ^ 3> 20 ft ^ 3$

8 0

4 years ago

Divide: (5 + 4i) / (-3 – 2i)

Phantasy [73]

Answer:

Step-by-step explanation:

$\frac{5+4i}{-3-2i}\\\\\mathrm{Apply\:complex\:arithmetic\:rule}:\\\\\quad \frac{a+bi}{c+di}\:=\:\frac{\left(c-di\right)\left(a+bi\right)}{\left(c-di\right)\left(c+di\right)}\:=\:\frac{\left(ac+bd\right)+\left(bc-ad\right)i}{c^2+d^2}\\\\a=5,\:b=4,\:c=-3,\:d=-2\\\\=\frac{\left(5\left(-3\right)+4\left(-2\right)\right)+\left(4\left(-3\right)-5\left(-2\right)\right)i}{\left(-3\right)^2+\left(-2\right)^2}\\\\Refine\\=\frac{-23-2i}{13}\\\\$