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3 years ago

3. Find the slope of a line passing through the points (-15, 48) (-3,60) *

Mathematics

2 answers:

Alisiya [41]3 years ago

6 0

The answer is 1
____________

Mars2501 [29]3 years ago

3 0

Answer:

slope (m) = 1

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Solve the equation. 31-12n=211

Mashutka [201]

Answer:

Step-by-step explanation:

31 - 12n = 21l Subtract 31 from both sides.

31-31-12n =211-31 Do the subtraction

-12n = 190 Divide by -12

-12n/-12 = 190/-12

n = -15.333333

6 0

3 years ago

What is the answer to this ?

harina [27]

Answer:

The answer is option 1.

Step-by-step explanation:

In order to solve x, you have to make x the subject, by multiplying each sides by 5, to get rid of 5 on the left side.

$\frac{x}{5} = 8$

$\frac{x}{5} \times 5 = 8 \times 5$

$x = 40$

5 0

3 years ago

In a pre algebra class containing 40 students, there are 18 freshmen, 19 sophomores, and 3 juniors. What fraction of the class a

dezoksy [38]

$\frac{9}{20}$ of the class consists of freshmen.

Step-by-step explanation:

Given,

Total number of students = 40

Number of freshmen = 18

Number of sophomores = 19

Number of juniors = 3

Fraction of freshmen = $\frac{Number\ of\ freshmen}{Total\ number\ of students}$

Fraction of freshmen = $\frac{18}{40}$

Fraction of freshmen = $\frac{9}{20}$

$\frac{9}{20}$ of the class consists of freshmen.

Keywords: fraction, division

Learn more about fractions at:

#LearnwithBrainly

7 0

3 years ago

????????????????????????????

frutty [35]

Answer:

Step-by-step explanation:

i took the same question hope this helps :D

5 0

3 years ago

Read 2 more answers

Find the extreme values of f(x y)=xy subject to the constraint x^2 + y^2 -4 = 0

KATRIN_1 [288]

Via Lagrange multipliers:

$L(x,y,\lambda)=xy+\lambda(x^2+y^2-4)$

$L_x=y+2\lambda x=0$
$L_y=x+2\lambda y=0$
$L_\lambda=x^2+y^2-4=0$

$yL_x=y^2+2\lambda xy=0$
$xL_y=x^2+2\lambda xy=0$
$\implies yL_x-xL_y=y^2-x^2=0\implies y^2=x^2$
$\implies x^2+y^2=4=2x^2\implies x^2=2\implies x=\pm\sqrt2\implies y=\pm\sqrt2$

So we have four critical points to consider, $(\sqrt2,\sqrt2),(-\sqrt2,\sqrt2),(\sqrt2,-\sqrt2),(-\sqrt2,-\sqrt2)$ . If both coordinates are positive or both are negative, we get a maximum value of $(\pm\sqrt2)^2=2$ ; otherwise, we get a minimum of $(-\sqrt2)(\sqrt2)=-2$ .

4 0

3 years ago