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kap26 [50]
3 years ago
8

What is the equation of the linear function represented by the table?

Mathematics
1 answer:
serg [7]3 years ago
3 0
It would be y = -x + 9
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A landscaper is creating a rectangular flower bed such that the width is half of the length.The area of the flower that is 34 ft
Leviafan [203]
The Area is A=l * w and we know that w=\frac{l}{2}

Therefore: 34=l * \frac{l}{2} = \frac{l^{2}}{2}
68=l^{2}
\sqrt{68}=l=8.246
and thus:
w=\frac{8.246}{2}
w=4.123
round to nearest tenth:
w=4.1feet
6 0
3 years ago
Read 2 more answers
A case of 8 items costs $9.44 at Costco. What is the price of each item?
Anarel [89]

Answer: $1.18 per item

Step-by-step explanation:

9.44/8=$1.18 per item

5 0
1 year ago
Find the distance between M(-2,3) and N(8,2)
AnnZ [28]

Answer:

d=\sqrt{101}

Step-by-step explanation:

Distance Formula: d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Simply plug in your 2 coordinates into the distance formula to find distance <em>d</em>:

d=\sqrt{(8-(-2))^2+(2-3)^2}

d=\sqrt{(8+2)^2+(-1)^2}

d=\sqrt{(10)^2+1}

d=\sqrt{100+1}

d=\sqrt{101}

6 0
3 years ago
Can someone please help me with my maths question​
DIA [1.3K]

Answer:

a. \  \dfrac{625 \cdot m}{27 \cdot n^{11}}

b. \  \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}

Step-by-step explanation:

The question relates with rules of indices

(a) The give expression is presented as follows;

\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}

By expanding the expression, we get;

\dfrac{m^3 \times n^{-8} \times 5^4 \times m^4}{\left 3^3 \times m^6 \times n^3}

Collecting like terms gives;

\dfrac{m^{(3 + 4 - 6)}  \times 5^4}{ 3^3 \times n^{3 + 8}} = \dfrac{625 \cdot m}{27 \cdot n^{11}}

\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}= \dfrac{625 \cdot m}{27 \cdot n^{11}}

(b) The given expression is presented as follows;

x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \div (x \cdot y^n)^4

Therefore, we get;

x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times  x^{-4} \times y^{-4 \cdot n}

Collecting like terms gives;

x^{3 \cdot m + 2 - 4} \times \left (y^{3 \cdot n - 3 -4 \cdot n}} \right ) = x^{3 \cdot m - 2} \times \left (y^{ - 3 -n}} \right ) = x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right )

x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right ) = \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}

x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times  x^{-4} \times y^{-4 \cdot n} =\dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}

4 0
3 years ago
Solve c =(a² + 3b)/4 for b.<br><br>Bruv take all my points I literally blow at math :'(​
Sati [7]

Answer:

b=\frac{4c-a^{2} }{3}

Step-by-step explanation:

First: multiply both sides by 4. 4 times c is 4c and the other sides cancels out as you are doing the inverse operation of division

Then, you have 4c=a^{2} +3b

Subtract both sides by a squared

DO NOT TAKE THE SQUARE ROOT! This is because it is a squared plus 3b so you have to do the inverse of addition

From that you get 4c-a^{2} =3b

Finally, divide both sides by 3.

You get  b=\frac{4c-a^{2} }{3}

4 0
3 years ago
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