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andrew-mc [135]
3 years ago
11

The quantity a varies directly with b and c and inversely with d. The quantity d is tripled. Which of the following must be true

for the relationship to remain the same? Both c and b must be tripled. The product of c and b must be tripled. Both c and b must be multiplied by 1/3. The product of c and b must be multiplied by 1/3.
Mathematics
1 answer:
Vinvika [58]3 years ago
8 0
Lets name the quantity Q
Q=k* \frac{b*c}{d} 

When quantity d is tripled, we get





Q=k \frac{b*c}{3d} 

To be the relationship the same, we see that we need to 
multiply  /numerator  / by  /3 


Q =  k \frac{3bc}{3d} 

So the answer should be 
"The product of c and b must be tripled."

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Caroline puts $400 into a savings account that earns 6% annually. Write a function that represents this situation, where t is th
DaniilM [7]

Answer: 19 years

Step-by-step explanation:

Given

Amount invested \$400

Rate of interest R=6\%

after 1-year amount becomes

A=400\left (1+6\%\right)\\A=400\left (1+0.06\right)\\A=400\times 1.06\\A=\$424

After n years it is

A=400\left (1.06\right)^n

Sum accumulated is \$1200

\Rightarrow 1200=400\left(1.06\right)^n\\\Rightarrow 3=\left(1.06\right)^n\\\text{Taking log both sides}\\\Rightarrow n=\dfrac{\ln 3}{\ln 1.06}\\\\\Rightarrow n=18.85\approx 19\ \text{years}

5 0
3 years ago
In the given figure, C is the midpoint of AB and M is the midpoint of AC . Find the length of NB if MC = 4 and AN = 14.
Elina [12.6K]

Answer:

Given MC = 4

AN = 14

To Find, the length of NB

Step-by-step explanation:

AB is a line which has midpoint “C”. Now the line is divided into two equal portion AC and CB.

The AC has midpoint “M” and MC is 4, so AM will also be 4.

N is the midpoint of CB. So, CB = CN + NB

Now we know AC = AM + MC = 4  + 4 =8

Given, AN = 14

AN = AC + CN

14 = 8 + CN

CN = 6

Since N is the midpoint of CB then, CN = NB

Therefore, the NB is 6

8 0
3 years ago
A rectangle has width that is 6 meters less than the length. The area of the rectangle is 280 square meters. Find the dimensions
salantis [7]

Dimensions are length 20 meter and width 14 meter

<em><u>Solution:</u></em>

Let "a" be the length of rectangle

Let "b" be the width of rectangle

Given that,

<em><u>A rectangle has width that is 6 meters less than the length</u></em>

Width = length - 6

b = a - 6

The area of the rectangle is 280 square meters

<em><u>The area of the rectangle is given by formula:</u></em>

Area = length \times width

<em><u>Substituting the values we get,</u></em>

Area = a \times (a-6)\\\\280 = a^2-6a\\\\a^2-6a -280=0

<em><u>Solve the above equation by quadratic formula</u></em>

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\\quad x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=1,\:b=-6,\:c=-280:\quad a_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:1\left(-280\right)}}{2\cdot \:1}

a =\frac{6 \pm \sqrt{36+1120}}{2}\\\\a = \frac{6 \pm \sqrt{1156}}{2}\\\\a = \frac{6 \pm 34}{2}\\\\Thus\ we\ have\ two\ solutions\\\\a = \frac{6+34}{2} \text{ or } a = \frac{6-34}{2}\\\\a = 20 \text{ or } a = -14

Since, length cannot be negative, ignore a = -14

<em><u>Thus solution of length is a = 20</u></em>

Therefore,

width = length - 6

width = 20 - 6 = 14

Thus dimensions are length 20 meter and width 14 meter

6 0
3 years ago
Combine and write the answer in descending power order (DPO):<br> 3y - y^3 + 2 -3y^2
dexar [7]

Answer:

im sorry

Step-by-step explanation:

ezxerfguibfyrxzwzx

8 0
3 years ago
9.
frozen [14]

Answer:

Your answer is C. Keith should plot -9 to the left of -8

because -9 <-8.

Step-by-step explanation:

I hope this helps lots. Plz mark brainliest

4 0
3 years ago
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