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3 years ago

I need answers please...

Mathematics

1 answer:

Lubov Fominskaja [6]3 years ago

7 0

Answer:

If you use the expression y=mx+b this will really help. M is the slope and B is the y intercept. For example, in question 1. the slope is M and M = 3 and the Y-intercept is b, b= -5

Step-by-step explanation:

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Ben made a huge number, writing out the consecutive natural numbers from 1 to 500:123 ... 10111213 ... 499500. Lusine erased fir

Aleksandr-060686 [28]

123 :D hope I helped

4 0

3 years ago

Please help ASAP! 10 points + BRAINLIEST!

worty [1.4K]

Hello good person. The answer is C, (4,7). If you replace the x's in both equations (4) and the y's i both equations (7) you would get your answer for both.

4 0

3 years ago

I’m not sure on what to do for these problems…can someone help me please

Ray Of Light [21]

Answer:

You need to find k, the constant of proportionality, by dividing y with its corresponding x.

3 0

3 years ago

I’m not good with geometry please help

jasenka [17]

Answer:

The value of t is 41

Step-by-step explanation:

The figure is 5 sided polygon.

The sum of interior angles of 'n' sided polygon is given by,

= $180(n-2) degrees$

here, n=5,

thus, $180(5-2) = 180(3) = 540 degrees$

All the given angles have linear pair in the polygon.

$(180-2t)+(180-3t)+(180-42)+(180-52)+(180-61) = 540$

$5t = 205$

Thus, t = 41

8 0

3 years ago

77. the volume of a cube is increasing at a rate of <img src="https://tex.z-dn.net/?f=10%20%5Cmathrm%7B~cm%7D%5E%7B3%7D%20%2F%20

Colt1911 [192]

Answer:

$\displaystyle \frac{4}{3}\text{cm}^2/\text{min}$

Step-by-step explanation:

Given

$\displaystyle \frac{dV}{dt}=10\:\text{cm}^3/\text{min}\\ \\V=s^3\\\\SA=6s^2\\\\\frac{d(SA)}{dt}=?}\:;s=30\text{cm}$

Solution

(1) Find the rate of the cube's edge length with respect to time at s=30:

$\displaystyle V=s^3\\\\\frac{dV}{dt}=3s^2\frac{ds}{dt}\\ \\10=3(30)^2\frac{ds}{dt}\\ \\10=3(900)\frac{ds}{dt}\\\\10=2700\frac{ds}{dt}\\\\\frac{10}{2700}=\frac{ds}{dt}\\\\\frac{ds}{dt}=\frac{1}{270}\text{cm}/\text{min}$

(2) Find the rate of the cube's surface area with respect to time at s=30:

$\displaystyle SA=6s^2\\\\\frac{d(SA)}{dt}=12s\frac{ds}{dt}\\ \\\frac{d(SA)}{dt}=12(30)\biggr(\frac{1}{270}\biggr)\\\\\frac{d(SA)}{dt}=\frac{360}{270}\biggr\\\\\frac{d(SA)}{dt}=\frac{4}{3}\text{cm}^2/\text{min}$

Therefore, the surface area increases when the length of an edge is 30 cm at a rate of $\displaystyle \frac{4}{3}\text{cm}^2/\text{min}$ .

6 0

2 years ago

I need answers please...​

I need answers please...