All categories

3 years ago

What is the measure of 2 G?

Mathematics

1 answer:

patriot [66]3 years ago

7 0

Answer:

28 =<G

Step-by-step explanation:

This is an isosceles triangle, so angle E and angle G must be the same.

<E = <G

28 =<G

You might be interested in

A small rectangular tissue box is 0.5 feet long, 0.2 feet wide, and 0.3 feet deep. The box is filled completely with tissues

Verizon [17]

Answer:

the volume of the tissue box would be 0.2×0.5×0.3. the volume of a rectangular prisim is V=l×w×h

6 0

2 years ago

Please help me! i will give 10 points and brainliest to those who show their work!!

Black_prince [1.1K]

Hopefully this helps.

8 0

3 years ago

Of five letters (a, b, c, d, and e), two letters are to be selected at random without replacement. how many possible selections

san4es73 [151]

Sent a picture of the solution to the problem (s). I sent the formula because I do not know if your calculator has the combination function. You will notice the denominator. I got it by the 2 picks you make and the 3 from what is left over. The 5 is from the total selection.

3 0

3 years ago

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$