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3 years ago

Graph the solution set for this inequality -6x-3y<-18

Mathematics

2 answers:

Vikentia [17]3 years ago

7 0

Answer:

Hope It Helps

Yuliya22 [10]3 years ago

5 0

Answer:

it is 27 because of my cacculations

Step-by-step explanation:

you would fist subtract your numbers and thne get rid of the 5

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How would a bigger down payment be beneficial to borrowers? a. A bigger down payment is only beneficial for a person applying fo

lawyer [7]

The correct option is D, A bigger down payment is money paid toward principal, interest-free, which also decreases the amount paid monthly.

<h3>What is interest?</h3>

The amount of money a lender or financial organization earns for lending out money is known as interest.

We know that monthly payments include the interest on the principal amount that is not been paid during the purchase of an item, if the down payment is more than the principal amount that the customer does not pay will decrease which will decrease the loan amount, thus, the interest on the principal amount will decrease as well.

Now, as the interest and the principal amount have both is been reduced, therefore, the monthly instalment(monthly payment) will be reduced as well.

Hence, the correct option is D, A bigger down payment is money paid toward principal, interest-free, which also decreases the amount paid monthly.

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3 years ago

PLEASE HELP ASAP MARK AS BRAINLIEST

MatroZZZ [7]

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4 0

3 years ago

A student is given that point P(a, b) lies on the terminal ray of angle Theta, which is between StartFraction 3 pi Over 2 EndFra

Harman [31]

Answer:

<em>The student made an error in step 3 because a is positive in Quadrant IV; therefore, </em>

<em /> $cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}$

Step-by-step explanation:

Given

$P\ (a,b)$

$r = \± \sqrt{(a)^2 + (b)^2}$

$cos\theta = \frac{-a}{\sqrt{a^2 + b^2}} = -\frac{\sqrt{a^2 + b^2}}{a^2 + b^2}$

Required

Where and which error did the student make

Given that the angle is in the 4th quadrant;

The value of r is positive, a is positive but b is negative;

Hence;

$r = \sqrt{(a)^2 + (b)^2}$

Since a belongs to the x axis and b belongs to the y axis;

$cos\theta$ is calculated as thus

$cos\theta = \frac{a}{r}$

Substitute $r = \sqrt{(a)^2 + (b)^2}$

$cos\theta = \frac{a}{\sqrt{(a)^2 + (b)^2}}$

$cos\theta = \frac{a}{\sqrt{a^2 + b^2}}$

Rationalize the denominator

$cos\theta = \frac{a}{\sqrt{a^2 + b^2}} * \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 + b^2}}$

$cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}$

So, from the list of given options;

<em>The student's mistake is that a is positive in quadrant iv and his error is in step 3</em>

3 0

3 years ago

Simplify this exponential function<br> b^ 8(2b)^5

xz_007 [3.2K]

Step-by-step explanation:

<h2>that's all</h2><h2>your answer</h2>

7 0

3 years ago

4x + 1/3 y 2 what is the value of the expression above when x = 2 and y = 3

GenaCL600 [577]

Hoi!

To solve this, first plug in the values for x and y.

x = 2, so anywhere you see x, put 2 in its place.

y = 3, so anywhere you see y, put 3 in its place.

$4(2) + \frac{1}{2}(3)$

4 × 2 = 8

$\frac{1}{2}$ × 3 = $1 \frac{1}{2}$

$8 +1 \frac{1}{2}$ = $9 \frac{1}{2}$

$9 \frac{1}{2}$ is your answer.

4 0

3 years ago

Read 2 more answers