All categories

3 years ago

Write two numbers that multiply to the value on top and add to the value on bottom.

Mathematics

1 answer:

vova2212 [387]3 years ago

7 0

Answer: -8 and 9
Explanation: -8 x 9 = -72 and -8 + 9 = 1

You might be interested in

Mr. Bridges paid $582.40 for a set of four tires for a truck about how much was the cost per tire

skelet666 [1.2K]

$582.40 / 4 tires = $145.60

Answer: $145.60/tile

5 0

3 years ago

What is 6.974 rounded to the nearest tenth

zloy xaker [14]

6.974 rounded to the nearest tenth is 7.0, because 7 is more than 5, and since 9's there, we make that 6 to a 7.

6 0

3 years ago

Read 2 more answers

Explain how to convert 6 feet into meters.

SCORPION-xisa [38]

Its 1.8288 meters use the conversion its equals this plz mark brainlest

7 0

3 years ago

Read 2 more answers

PLEASE HELP I HAVE 5 MINUTES LEFT I WILL REPORT IF YOU GIVE A LINK OR UNHELPFUL ANSWER I WILL GIVE BRAINILEST 100 POINTS

Lorico [155]

Answer:

7-b≤c

7-b≥c

Step-by-step explanation:

I am so sorry if it is wrong.... I tried

7 0

3 years ago

The 6-lb particle is subjected to the action of its weight and forces F1 = 52i + 6j - 2tk6 lb, F2 = 5t 2 i - 4tj - 1k6 lb, and F

olga2289 [7]

Answer:

r=294.9m

Step-by-step explanation:

The forces on the particle are

$W=mg\hat{j}\\F_{1}=52\hat{i}+6\hat{j}-2t\hat{k}\\F_{2}=5t^{2}\hat{i}-4t\hat{j}-1\hat{k}\\F_{3}=(5-2t)\hat{i}$

Now , we sum all these forces to get the net force

$F_{T}=W+F_{1}+F_{2}+F_{3}\\F_{T}=(52+5t^{2}+5-2t)\hat{i}+((6+6-4t)\hat{j}+(-2t-1)\hat{k}\\F_{T}=(57-2t+5t^{2})\hat{i}+(12-4t)\hat{j}+(-2t-1)\hat{k}\\$

we can use the fact F=m*a and integrate the acceleration

$a(t)=\frac{1}{m}F(t)\\\\v(t)=\int a(t)dt=\frac{1}{m}\int{F_{T}}dt\\\\v(t)=\frac{1}{m}[(57t-t^{2}+\frac{5}{3}t^{3})\hat{i}+(12t-2t^{2})\hat{j}+(-t^{2}-t)\hat{k}]\\\\r(t)=\int v(t)dt=\frac{1}{m}[(\frac{57}{2}t^{2}-\frac{1}{3}t^{3}}+\frac{5}{4}t^{4})\hat{i}+(6t^{2}-\frac{2}{3}t^{3})\hat{j}+(-\frac{1}{3}t^{3}-\frac{1}{2}t^{2})]$

and we evaluate in r(2) an we take the norm to obtain the distance

$r(2)=\frac{1}{m}[\frac{394}{3}\hat{i}+\frac{56}{3}\hat{j}-\frac{14}{3}\hat{k}]\\|r(2)|=\frac{1}{m}\sqrt{[(\frac{394}{3})^{2}+(\frac{56}{3})^{2}+(\frac{14}{3})^{2}]}\\|r(2)|=\frac{132.73}{0.45}=294.9m$

I hope this is useful for you

regards

8 0

4 years ago