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3 years ago

What is the equation of the line that passes through the point (4,-3)(and has a slope of 11?

Mathematics

1 answer:

emmasim [6.3K]3 years ago

6 0

Answer: -3 = 11(4) + b

Step-by-step explanation: y = mx + b, -3 is the y, 11 is the slope (m), 4 is the x, b is the y-intercept which is unknown unless you solve for it.

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The product of four consecutive odd numbers(all four positive or all four negative)is 62985.find the two possible sets of four n

emmasim [6.3K]

Answer:

Step-by-step explanation:

Let X be the first number.

(X)*(X+2)*(X+4)*(X+6) = 62985

(X^2 + 6X)^2 + 8(X^2 + 6X) = 62985

X = 13, AND X= -19

A. X = 13

(13)(15(17)(19) = 62985

B. X = -19

(-19)(-17)(-15)(-13) = 62985

8 0

2 years ago

How does a set of integers differs from the set of whole numbers ? Please help

MariettaO [177]

The difference between set of whole numbers and a set of integers is that:

Whole numbers = {0, 1, 2, 3, 4, 5...}
Integers = {..., -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5...}

If we name contents of the first set as n, in the second set you can have 2n-1 numbers (because we count 0 only once in both sets).

7 0

3 years ago

2. Dr. Scott bought 3 pound of oranges for $1.50. How much would 7

n200080 [17]

Do 1.50 divided by 3 to get the unit price.

6 0

3 years ago

Use the table to solve the inequality x−5≥7.

miskamm [114]

Answer: x $\geq$ 12

Step-by-step explanation:

6 0

2 years ago

I need help w/ this!! thank you

Fittoniya [83]

<h3>Answer:</h3>

$\boxed{\pink{\sf \leadsto Yes \ there \ is \ a \ solution \ of \ the \ given \ inequality .}}$

<h3>Step-by-step explanation:</h3>

A inequality is given to us and we need to convert it into standard form and see whether if it has a solution . So let's solve the inequality.

The inequality given to us is :-

$\bf\implies |2y + 3 | - 1 \leq 0 \\\\\bf\implies |2y+3|\leq 1 \\\\\bf\implies (|2y+3|)^2 \leq 1^2 \\\\\bf\implies (2y+3)^2 \leq 1 \\\\\bf\implies (2y)^2+3^2+2(2y)(3) \leq 1 \\\\\bf\implies 4y^2+9+12y - 1 \leq 0 \\\\\bf\implies 4y^2+12y+8 \leq 0 \\\\\bf\implies 4( y^2 + 3y + 2 ) \leq 0 \\\\\bf\implies y^2+3y +2 \leq 0 \:\:\bigg\lgroup \purple{\bf Standard \ form \ of \ inequality }\bigg\rgroup \\\\\bf\implies y^2y+2y+y+2 \leq 0 \\\\\bf\implies y(y+2)+1(y+2)\leq 0 \\\\\bf\implies ( y+2)(y+1)\leq 0 \\\\\bf\implies \boxed{\red{\bf y \leq (-2) , (-1) }}$

Let's plot a graph to see its interval . Graph attached in attachment .

Now we can see that the Interval notation of would be ,

$\boxed{\boxed{\orange \tt \purple{\leadsto}y \in [-2,-1] }}$

<h3>Hence the standard form of inequality is y²+3y +2 ≤ 0 and the Solution set of the inequality is [ -2 , -1 ] .</h3>

4 0

3 years ago