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Norma-Jean [14]

1 year ago

12

jack works after school. each day he is paid a set amount, plus an hourly wage. look at the table. choose linear function f jack

can use to find his pay.

1 answer:

Vilka [71]1 year ago

4 0

Answer:

C

Step-by-step explanation:

I found the slope which is 28-22 / 1.5-1 = 12

I plug in the hours to 12x+10 and it matched

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A 2-digit number with x units and 7 less units than tens.

Hatshy [7]

Each two digit number has two numbers (duh!). Let's allow the tens digit to be x and the units digit to be y. Tens digit is 3 less than the units digit: x = y-3 Original number is 6 more than 4 times the sum of the digits: 10x+y-6 = 4x + 4y This gives us simultaneous equations!First let's clear the mess: 1. x= y-32. 6x-3y=6 Substitute 1 into 2: 6(y-3) -3y =66y - 18 - 3y = 6
3y = 24y = 8 Our units digit is 8 Substitute y= 8 into 1. x = y - 3x = 5 Our tens digit is 5 Therefore, our number is 58

7 0

2 years ago

7a = 28 Is a=21 a solution?<br><br> yes or no

bagirrra123 [75]

No, because 7 x 21 is 147.

4 0

1 year ago

Read 2 more answers

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

2 years ago

The angular velocity of a point on a circle with radius of 12 feet is 15 radians per second. Find the speed of the point on the

Lyrx [107]

Angular speed:
(24*pi)*(15 pi/2pi) ft/sec:
565 ft/sec

5 0

2 years ago

Read 2 more answers

NeX [460]

Answer/explanation

4(3x+5)=-4

You will distribute the four to every term in the parenthesis,

12x+20=-4

Then you will combine like terms

12x+20=-4

-20 -20

12x=-24

Then you will divide 12 to isolate x

$\frac{12x}{-24} =\frac{-24}{-24}$

Obviously, -24 cancels out and you would get x=-2

4 0

2 years ago