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2 years ago

7

What is the slope of the line in the graph?

1 answer:

Goshia [24]2 years ago

7 0

Answer:

My answer would be y=2x+1

Step-by-step explanation:

You might be interested in

1/3 divided by 2 divided

KatRina [158]

Answer:

1/6

Step-by-step explanation:

1/3 diveded by 2/1

1/3x1/2 = 1/6

Hope it helps:)

6 0

1 year ago

A rectangle has dimensions 3x-1 and x+6. Write an expression for the area of the rectangle as a polynomial in standard form

WITCHER [35]

Answer:

the area product is simply (3x+5).(x+7 which can multiply if desired to obtain highlight3x/2+ 26x + 35.

You can use Completion of the Square on the trinomial product to put this trinomial into standard form. You would want this form to be like (x-h)2 +k.

7 0

2 years ago

The line l cuts the y-axis at (0, 5). l also passes through the point (2, 1). find the equation of the line which is parallel to

Over [174]

Answer: $y=-2x+6$

Step-by-step explanation:

The slope of line l is

$\frac{5-1}{0-2}=-2$

Since parallel lines have the same slope, the slope of the line we need to find is -2.

Substituting into point-slope form,

$y=-2(x-3)\\\\y=-2x+6$

3 0

1 year ago

7n + 5 =68.... how to solve this problem and show my work

Evgen [1.6K]

Answer:

n = 9

Step-by-step explanation:

7n + 5 = 68

7n + 5 - 5 = 68 - 5

7n = 63

7n / 7 = 63 / 7

n = 9

5 0

2 years ago

Express the number 220 as the sum of four numbers that form a geometric progression such that the third term is greater than the

siniylev [52]

Answer:

Step-by-step explanation:

The first four terms of geometric series is:

$a,ar,ar^2,ar^3$

Since, we have given information that the third number is greater than 44 that means:

$ar^2=a+44$

Above equation can be rewritten as:

$a(r^2-1)=44$

Now, using:

$a^2-b^2=(a+b)(a-b)$

Here, a=r,b=1

$a(r+1)(r-1)=44$ (1)

The sum of first four terms is:

$a+ar+ar^2+ar^3=220$

$a(1+r+r^2+r^3)=220$

$a(1(1+r)+r^2(1+r))=220$

$a((1+r)(1+r^2))=220$ (2)

Divide equation (2) by (1) we get:

$\frac{r^2+1}{r-1}=\frac{220}{44}$

$r^2+1=5r-5$

$\Rightarrow r^2-5r+6=0$

$\Rightarrow r^2-3r-2r+6=0$

$\Rightarrow r(r-3)-2(r-3)=0$

$\Rightarrow (r-2)(r-3)=0$

$\Rightarrow r=2,3$

CASE1: When r=2 in $ar^2=a+44$

$a(4)=a+44$

$3a=44$

$a=\frac{44}{3}$

CASE2:When r=3 in $ar^2=a+44$

$a(3)^2=a+44$

$\Rightarrow 9a=a+44$

$\Rightarrow 8a=44$

$\Rightarrow a=\frac{44}{8}=\frac{11}{2}$

The series becomes:

From CASE1: $\frac{44}{3},\frac{44\cdot 2}{3},\frac{44\cdot 2^2}{3},\frac{44\cdot 2^3}{3}$

$\Rightarrow \frac{44}{3},\frac{88}{3},\frac{176}{3},\frac{352}{3}$

From CASE2: $\frac{11}{2},\frac{11\cdot 3}{2},\frac{11\cdot 3^2}{2},\frac{11\cdot 3^3}{2}$

$\Rightarrow \frac{11}{2},\frac{33}{2},\frac{99}{2},\frac{297}{2}$

6 0

2 years ago

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