2/3x+5=21 pls use shoe work

To create a giant gemstone, sara first made two identical square pyramids that each had a base area of 100 square inches. Then s

inysia [295]

Answer:

8 inches.

Step-by-step explanation:

From the statement we have that they first made two identical square pyramids, each with a base area of 100 square inches.

Ab = s ^ 2 = 100

Therefore each side would be:

s = (100) ^ (1/2)

s = 10

So, side of the square base = 10 inches

Then they tell us that they glued the bases of the pyramids together to form the precious stone. The surface area of the gemstone is 520 square inches, so for a single pyramid it would be:

Ap = 520/2 = 260

For an area of the square pyramid we have the following equation:

Ap = 2 * x * s + s ^ 2

Where x is the height of each triangular surface and s is the side of the square base

Replacing we have:

260 = 2 * x * 10 + 10 ^ 2

20 * x + 100 = 260

20 * x = 160

x = 160/20

x = 8

Therefore, the value of x is 8 inches.

5 0

3 years ago

Which graph or graphs appear to show a sinusoid ?

Mnenie [13.5K]

Answer:

D

Step-by-step explanation:

Sinusoid graphs look like regular waves.

It is obvious that only I and III look like waves.

II is known as a stepwise function.

5 0

3 years ago

Read 2 more answers

Which equation can be simplified to find the inverse of y = x2 – 7?

iragen [17]

The last one is right

3 0

3 years ago

Plz help the ones with question marks

3241004551 [841]

Answer:

1(5/16)=1.3125

7(5/16)=7.3125

26(5/16)=26.3125

Step-by-step explanation:

Given that 5/16=0.3125, the mixed fraction 1(5/16)=1+(5/16)=1+0.3125=1.3125

Also, 7(5/16)=7+(5/16)=7+0.3125=7.3125

Furthermore,

26(5/16)=26+(5/16)=26+0.3125=26.3125

In the same vein, given that 4/15=0.266

1(4/15)=1+(4/15)=1+0.266=1.266

17(4/15)=17+(4/15)=17+0.266=17.266

23(4/15)=23+(4/15)=23+0.266=23.266

7 0

3 years ago

Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat

eimsori [14]

Answer:

There are NO real roots for this equation. The only roots have imaginary parts and therefore cannot be represented on the real x-axis.

Step-by-step explanation:

We notice that the expression on the left of the equation is a quadratic with leading term $2x^2$ , which means that its graph is that of a parabola with branches going up.

Therefore, there can be three different situations:

1) if its vertex is ON the x axis, there would be one unique real solution (root) to the equation.

2) if its vertex is below the x-axis, the parabola's branches are forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will have NO real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently.

We recall that the x-position of the vertex for a quadratic function of the form $f(x)=ax^2+bx+c$ is given by the expression:

$x_v=\frac{-b}{2a}$

Since in our case $a=2$ and $b=-3$ , we get that the x-position of the vertex is:

$x_v=\frac{-b}{2a}\\x_v=\frac{-(-3)}{2(2)}\\x_v=\frac{3}{4}$

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = 3/4:

$y_v=f(\frac{3}{4})=2( \frac{3}{4})^2-3(\frac{3}{4})+4\\f(\frac{3}{4})=2( \frac{9}{16})-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{18}{8}+\frac{32}{8}\\f(\frac{3}{4})=\frac{23}{8}$

This is a positive value for y, therefore we are in the situation where there is NO x-axis crossing of the parabola's graph, and therefore no real roots.

We can though estimate a few more points of the parabola's graph in order to complete the graph as requested in the problem. For such we select a couple of x-values to the right of the vertex, and a couple to the right so we can draw the branches. For example: x = 1, and x = 2 to the right; and x = 0 and x = -1 to the left of the vertex:

$f(-1) = 2(-1)^2-3(-1)+4= 2+3+4=9\\f(0)=2(0)^2-3(0)+4=0+0+4=4\\f(1)=2(1)^2-3(-1)+4=2-3+4=3\\f(2)=2(2)^2-3(2)+4=8-6+4=6$

See the graph produced in the attached image.

4 0

3 years ago