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3 years ago

7

Roberto wants to reduce a drawing that is 12 inches long by 9 inches wide. If his new drawing is 8 inches long, how wide will it

be?

2 answers:

Bess [88]3 years ago

6 0

Answer: It would be 6 inches wide.

Step-by-step explanation: Well two thirds of 12 in. is 8in. So two thirds of 9in. would be 6 inches.

erastova [34]3 years ago

5 0

Answer 67

Step-by-step explanation:

u have to divide 9 by 8 and then multiply 12 and then u get 67 ;) i hope this help

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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:

<u>Given</u>

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

<u>Solution</u>

(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

You have 4 different colored t-shirt (white, blue, red, and yellow) and 3 different colored

Vesna [10]

2/7 because you have 2 chances to pull blue and 7 items

4 0

3 years ago

Read 2 more answers

Help please!!!!!!!!!!!!!!!!!!!!!

chubhunter [2.5K]

Answer:

1. 130° 2. 128°

3. 66° 4. 100°

5. 90° 6. 81°

7. 53° 8. 58°

Step-by-step explanation:

8 0

2 years ago

Andrej [43]

Answer:

16

Step-by-step explanation:

Given that :

r + 9 = 4 greater Than s

r - 1 1 is how much less than s

r + 9 = s + 4

Solve for r

Subtract 9 from both sides

r + 9 - 9 = s + 4 - 9

r = s - 5

Hence,

r - 11 = (s - 5) - 11

r - 11 = s - 5 - 11

r - 11 = s - 16

r - 11 = s - 16

Therefore, r - 11 is 16 less than s

4 0

2 years ago

A sequence of three real numbers forms an arithmetic progression with a ﬁrst term of 9. If 2 is added to the second term and 20

Alexxandr [17]

Answer:

The smallest possible value for the third term of the geometric progression is 1.

Step-by-step explanation:

Given : A sequence of three real numbers forms an arithmetic progression with a first term of 9. If 2 is added to the second term and 20 is added to the third term, the three resulting numbers form a geometric progression.

To find : What is the smallest possible value for the third term of the geometric progression?

Solution :

The arithmetic progression is given by $a,a+d,a+2d$

First term a=9

Second term - 2 is added to second term i.e. $a+d+2=9+d+2=11+d$

Third term - 20 is added to the third term i.e. $a+2d+20=9+2d+20=29+2d$

The geometric progression is given by $a,ar,ar^2$

First term a=9

Second term - $ar=11+d$

Third term $ar^2=29+2d$

r is the common ratio which is second term divided by first term,

So, $r=\frac{ar}{a}=\frac{11+d}{9}$

or third term divided by second term,

So, $r=\frac{ar^2}{ar}=\frac{29+2d}{11+d}$

Equating both the r,

$\frac{11+d}{9}=\frac{29+2d}{11+d}$

Cross multiply,

$(11+d)\times (11+d)=(29+2d)(9)$

$11^2+11d+11d+d^2=261+18d$

$121+22d+d^2-261-18d=0$

$d^2+4d-140=0$

Solving by middle term split,

$d^2+14d-10d-140=0$

$d(d+14)-10(d+14)=0$

$(d+14)(d-10)=0$

$d=-14,10$

Substituting the value of d in the third term,

Third term $ar^2=29+2d$

When d=-14,

Third term $ar^2=29+2(-14)=29-28=1$

When d=10,

Third term $ar^2=29+2(10)=29+20=49$

Therefore, The smallest possible value for the third term of the geometric progression is 1.

7 0

3 years ago