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3 years ago

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;

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