Find the measure of "theta". Round all answers to the nearest tenth.

Solve for x. 47x=2747x=27

Svetach [21]

Divide by the coefficient of x.
x = 27/47

7 0

2 years ago

Please help me on this one

NikAS [45]

Answer:

55.4 in³

Step-by-step explanation:

The formula for the volume of a pyramid can be found:

V = $\frac{1}{3}Bh$ , where B = area of the base and h = height of the pyramid

Since the base is a triangle, we need to use the formula for the area of a triangle:

A = $\frac{1}{2}bh$ = $\frac{1}{2}(5.7)(6.2)=17.67in^{2}$

Next, multiply the area of the base by the height of the pyramid:

17.67 x 9.4 = 166.098 ÷ 3 ≈ 55.4 in³

3 0

3 years ago

1 1/8 +5/6 =<br><br> And then subtract 1/2 from the answer

kari74 [83]

Alright so start of with solving 1 1/8+5/6

Solve and you get 1 23/24

Now to subtract you have your answer subtracted by 1/2:

1 23/24-1/2=1 11/24

And so you get 1 11/24

~Evie

8 0

3 years ago

The graph h = −16t^2 + 25t + 5 models the height and time of a ball that was thrown off of a building where h is the height in f

Thepotemich [5.8K]

Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

h ----> the height of a ball in feet

t ---> the time in seconds

we have

$h(t)=-16t^{2}+25t+5$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

so

The x-coordinate of the vertex represent the time when the ball reach the maximum

Find the vertex

Convert the equation in vertex form

Factor -16

$h(t)=-16(t^{2}-\frac{25}{16}t)+5$

Complete the square

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}$

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}$

Rewrite as perfect squares

$h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

The vertex is the point $(\frac{25}{32},\frac{945}{64})$

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

$0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

$16(t-\frac{25}{32})^{2}=\frac{945}{64}$

$(t-\frac{25}{32})^{2}=\frac{945}{1,024}$

square root both sides

$(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}$

$t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}$

the positive value is

$t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec$

8 0

3 years ago

Which EST graph correctly represents the information on the table

garik1379 [7]

Answer:

the first one

Step-by-step explanation:

A graph showing the Earliest Start Times (EST) for project tasks is computed left to right based on the predecessor task durations. For dependent tasks, the earliest start time will be the latest of the finish times of predecessor tasks.

The first graph appears to appropriately represent the table values, using edges to represent task duration, and bubble numbers to represent start times.

The second graph does not appropriately account for duration of predecessor tasks.

The third graph seems to incorrectly compute task completion times (even if you assume that the edge/bubble number swap is acceptable).

3 0

3 years ago