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andreyandreev [35.5K]

3 years ago

10

Which is an equation of the line with a slope of 2 that passes through (3,−5)?

1 answer:

Naya [18.7K]3 years ago

7 0

Answer:

equation of a line= y-y¹=m(x-x¹)

y--5=2(x-3)

y+5=2x-6

y=2x-6+5

y=2x-1

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Which values are solutions to the inequality below? √x ≤ 5

Romashka [77]

$\sqrt{x} \leq 5$

Find the real region for $\sqrt{x}$

$\sqrt{x}$ is real for $x \geq 0$

$= x \geq 0$

$\sqrt{x} \leq 5$

Square both sides:

$( \sqrt{x} )^2 \leq 5^2$

$5^2 = 5*5 =25$

Refine:

$x \leq 25$

Answer D

hope this helps!

5 0

3 years ago

For which number is this the expanded form? 9 × 10 + 2 × 1 + 3 × (1/10) + 8 × (1/100) Group of answer choices 98.08 93.48 9.238

photoshop1234 [79]

Answer:

Well it is i think

Step-by-step explanation:

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6 0

2 years ago

Please help me with this! Picture is provided and only answer is you know it!

Sophie [7]

Answer:

Option A

Step-by-step explanation:

Number of components assembled by the new employee per day,

N(t) = $\frac{50t}{t+4}$

Number of components assembled by the experienced employee per day,

E(t) = $\frac{70t}{t+3}$

Difference in number of components assembled per day by experienced ane new employee

D(t)= E(t) - N(t)

D(t) = $\frac{70t}{t+3}-\frac{50t}{t+4}$

= $\frac{70t(t+4)-50t(t+3)}{(t+3)(t+4)}$

= $\frac{70t^2+280t-50t^2-150t}{(t+3)(t+4)}$

= $\frac{20t^2+130t}{(t+3)(t+4)}$

= $\frac{10t(2t+13)}{(t+4)(t+3)}$

Therefore, Option A will be the answer.

3 0

3 years ago

Amy is 11 ⅚ years old. Camden is 1 ⅓ years younger than amy and Jane is 1 ¼ years younger than Camden How old is Jane

erastova [34]

Answer:

Jane is 9 1/4 years old

Step-by-step explanation:

Camden is 10 1/2 years old and Jane is (10 2/4 - 1 1/4) which equals 9 1/4 years

5 0

3 years ago

Find the equation of line b described below, in slope-intercept form

Serjik [45]

Given:

Line a is perpendicular to line b .

Line a passes through the points (1,-8) and (9,-12) .

Line b passes through the point (-8, -16).

To find:

The equation of b.

Solution:

Line a passes through the points (1,-8) and (9,-12) . So, slope of line a is

$m=\dfrac{y_2-y_1}{x_2-x_1}$

$m_a=\dfrac{-12-(-8)}{9-1}$

$m_a=\dfrac{-12+8}{8}$

$m_a=\dfrac{-4}{8}$

$m_a=-\dfrac{1}{2}$

Product of slopes of two perpendicular lines is -1.

$m_a\times a_b=-1$

$-\dfrac{1}{2}\times a_b=-1$

$b=2$

Slope of line b is 2.

If a line passing through a point $(x_1,y_1)$ with slope m, then equation of line is

$y-y_1=m(x-x_1)$

Line b passing through (-8,-16) with slope 2. So, equation of line b is

$y-(-16)=2(x-(-8))$

$y+16=2(x+8)$

$y+16=2x+16$

Subtract 16 from both sides.

$y=2x$

Therefore, the equation of line b is $y=2x$ .

7 0

3 years ago