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4 years ago

How do you solve #4, show your work...

Mathematics

1 answer:

Thepotemich [5.8K]4 years ago

4 0

6-3=3

(6+3)2=81

3/81=.037 (repeating)

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How can i solve 5×23×2

VARVARA [1.3K]

multiply 5 and 2 then that with 23.

5x2=10 23x10=230

3 0

3 years ago

Read 2 more answers

A parking garage charges 21.00 for 6 hours how much does it charge per hour

alexdok [17]

Answer:

If a parking garage charges $21 for 6 hours in order to find out how much it charges per hour you will need to divide 21 by 6:

21÷6=3.50

So in conclusion a parking garage charges

$3.50/hour

5 0

3 years ago

We are given that lines a and c intersect at point S. Translate line a down line c until point S reaches point Q. Call the new l

Pavel [41]

Answer:

The lines a and b are parallel

Step-by-step explanation:

When we will translate line a, then the new line b will be formed.

Now, as translation preserves orientation each point on line a will be equidistant from line b , thus holding the condition and satisfying the property of two parallel lines.

Hence, line a and line b are parallel to each other.

8 0

4 years ago

Read 2 more answers

.) A florist wishes to make bouquets of mixed spring flowers. Each bouquet is to be made up of tulips at $10 a bunch and daffodi

LekaFEV [45]

We have the following:

let x a tulips

let y daffodils

therefore:

$\begin{gathered} x+y=15\rightarrow x=15-y \\ 10x+7y=(x+y)8\rightarrow10x+7y=120 \end{gathered}$

solving:

$\begin{gathered} 10\cdot(15-y)+7y=120 \\ 150-10y+7y=120 \\ 10y-7y=150-120 \\ 3y=30 \\ y=\frac{30}{3}=10 \end{gathered}$

for x:

$x=15-10=5$

Therefore, the answer is 5 tulips and 10 daffodils

let x a 30% silver

let y a 55% silver

$\begin{gathered} x+y=800\rightarrow x=800-y \\ 30x+55y=40\cdot(x+y)\rightarrow30x+55y=32000 \end{gathered}$

solving:

$\begin{gathered} 30\cdot(800-y)+55y=32000 \\ 24000-30y+55y=32000 \\ 25y=32000-24000 \\ y=\frac{8000}{25} \\ y=320 \end{gathered}$

for x:

$x=800-320=480$

Therefore, the answer is 480 pounds of 30% silver and 320 pounds of 55% silver

3 0

1 year ago

Yuri thinks that 3/4 is a root of the following function.

sineoko [7]

Given:

The polynomial function is

$q(x)=6x^3+19x^2-15x-28$

Yuri thinks that $\dfrac{3}{4}$ is a root of the given function.

To find:

Why $\dfrac{3}{4}$ cannot be a root?

Solution:

We have,

$q(x)=6x^3+19x^2-15x-28$

If $\dfrac{3}{4}$ is a root, then the value of the function at $\dfrac{3}{4}$ is 0.

Putting $x=\dfrac{3}{4}$ in the given function, we get

$q(\dfrac{3}{4})=6(\dfrac{3}{4})^3+19(\dfrac{3}{4})^2-15(\dfrac{3}{4})-28$

$q(\dfrac{3}{4})=6(\dfrac{27}{64})+19(\dfrac{9}{16})-\dfrac{45}{4}-28$

$q(\dfrac{3}{4})=3(\dfrac{27}{32})+\dfrac{171}{16}-\dfrac{45}{4}-28$

$q(\dfrac{3}{4})=\dfrac{81}{32}+\dfrac{171}{16}-\dfrac{45}{4}-28$

Taking LCM, we get

$q(\dfrac{3}{4})=\dfrac{81+342-360-896}{32}$

$q(\dfrac{3}{4})=\dfrac{-833}{32}\neq 0$

Since the value of the function at $\dfrac{3}{4}$ is not equal to 0, therefore, $\dfrac{3}{4}$ is not a root of the given function.

4 0

3 years ago