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3 years ago

Which line has a y-intercept of –2?a)

Mathematics

1 answer:

Elden [556K]3 years ago

8 0

Answer:

Step-by-step explanation:

It passes through -2 on the y-axis

HOPE THIS HELPS! (:

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Solve each inequality, graph the solution on the number line, and write the solution in interval notation

Lubov Fominskaja [6]

Answer:

$(-\infty,+\infty)$

Step-by-step explanation:

The given inequality is 3x-2>4 or $5x-3\leq 7$

We group similar terms to obtain: 3x>4+2 or $5x\le7+3$

Simplify to get:

3x>6 or $5x\le10$

x>2 or $x\le 2$

This implies that the solution set is all real numbers.

The solution in interval notation is $(-\infty,+\infty)$

5 0

3 years ago

Use front-end estimation to estimate the sum. 56.7 + 23.44

Reika [66]

57+23 = 80
not rounded it's 80.14

7 0

3 years ago

Angle DEF measures 75 . What is the measure of angle CfD?

vladimir1956 [14]

Any diagram provided!??

3 0

3 years ago

a buret is a tool designed to transfer precise amounts of liquid .A buret initially contains 70.00 millimeters of a solution and

katrin [286]

Let:

Vbu= Volume of the buret

Vbk= Volume of the beaker

A buret initially contains 70.00 millimeters of a solution and a beaker initially contains 20.00 ml of the solution the buret drips solution into the Beaker. each drip contains 0.05 mL of solution, therefore:

x = Number of drips

a = volume of each drip

$\begin{gathered} Vbu=70-ax \\ Vbk=20+ax \\ \text{where:} \\ a=0.05 \\ Vbu=70-0.05x \\ Vbk=20+0.05x \end{gathered}$

after how many drips will the volume of the solution in the buret and beaker be equal ? Vbu = Vbk:

$\begin{gathered} Vbu=Vbk \\ 70-0.05x=20+0.05x \\ \text{Solve for x:} \\ 0.1x=70-20 \\ 0.1x=50 \\ x=\frac{50}{0.1} \\ x=500 \end{gathered}$

5 0

1 year ago

Find the x-intercepts of the parabola with vertex (-4,2) and y-intercept (0,-30). Write your answer in this form: (x1, y1),(x2,

Ainat [17]

Vertex form of a parabola
y = a (x - h)^2 + k 

where (h, k) is the vertex 
Substituting the values of h and k.
we get,

y = a(x + 4)^2 + 2 

substituting in the point (0, -30) for x and y
-30 = a (0 + 4)^2 + 2

solve for a,

-30 = 16 a + 2 
-32 = 16 a 
-2 = a 

y = -2(x + 4)^2 + 2 

Put y = 0 

-2 x^2 - 16 x - 30 = 0 
-2(x^2 + 8 x + 15) = 0 
x^2 + 8 x + 15 = 0 
(x + 3)(x + 5) = 0 

x = -3
x = -5

3 0

3 years ago