All categories

3 years ago

What is 6 1/2 + 9 3/8

Mathematics

2 answers:

san4es73 [151]3 years ago

4 0

Answer:

the answer is 15 7/8

Tatiana [17]2 years ago

4 0

Answer: 15 7/8

Step-by-step explanation:

Add the whole numbers first 6+9 to getting 15

Now find the lowest common denominator for both the fractions. In this case it is 8.

3/8 would stay the same however 1/2 would become 4/8. 4/8 + 3/8 is 7/8.

15 add 7/8 is 15 7/8

You might be interested in

Distance: (-2,7) and (2,-5)

scZoUnD [109]

Does it say anything about slope?

6 0

3 years ago

Applying the properties of exponents, which mathematical operation changes 2 1/3 to 2?

e-lub [12.9K]

Hope this helps!

Stay safe and have a great day :D

6 0

3 years ago

Please tell me how o do this

larisa86 [58]

Answer:

you might have to use an app that scans that specific problem :)

Step-by-step explanation:

5 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:

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