All categories

3 years ago

Which lines have a y-intercept at (0,4)?

Mathematics

2 answers:

Sati [7]3 years ago

8 0

Answer:

y=-3x+4

Step-by-step explanation:

Using the formula y=mx+b

mx= slope b= y-intercept

hence,

y=-3x+4 has a y-intercept of (0,4)

Sergeu [11.5K]3 years ago

6 0

Answer:

A y=3x +4

Step-by-step explanation:

since when x=0 then y will be at +4

Y=3(0)+

You might be interested in

When you solve a problem involving money,what can a negative answer represents

AveGali [126]

Maybe the money you owe someone

4 0

2 years ago

Three pizzas are shared equally among 12 people.what fraction of pizza will each person get

8_murik_8 [283]

Answer:

1/4

Step-by-step explanation:

3/12 =0.25

0.25 = 1/4

8 0

2 years ago

Read 2 more answers

Can someone please help me with this

lana [24]

Answer:

3x-10+70°=180°(Supplementary angles)

3x+60°=180°

3x=180°-60°

3x=120°

3x/3 =120°/3

x=40°

Step-by-step explanation:

Hope that this is helpful.

Have a wonderful day.

6 0

2 years ago

12. Determine the area of the shaded region. 12 in 12 in

vekshin1

Answer:

216/7 in²

Step-by-step explanation:

Area of shaded region = Area of square - Area of circles
Radius of circle = 3 in
Area of circle = πr²
Area of square = s²

<u>Solution:</u>

Area of shaded region = Area of square - Area of circles
=> Area of shaded region = (12²) - 4(22/7 x 3 x 3)
=> Area of shaded region = (144) - 4(22/7 x 9)
=> Area of shaded region = (144) - 4(198/7)
=> Area of shaded region = 144 - 792/7
=> Area of shaded region = 144 x 7/7 - 792/7
=> Area of shaded region = 1008/7 - 792/7
=> Area of shaded region = 1008/7 - 792/7
=> Area of shaded region = 216/7 in²

3 0

2 years ago

Maths functions question!!

Marina86 [1]

Answer:

5) DE = 7 units and DF = 4 units

6) ST = 8 units

$\textsf{7)} \quad \sf OM=\dfrac{3}{2}\:units$

8) x ≤ -3 and x ≥ 3

Step-by-step explanation:

<u>Information from Parts 1-4:</u>

brainly.com/question/28193969

$f(x)=-x+3$

$g(x)=x^2-9$

A = (3, 0) and C = (-3, 0)

Points A and D are the <u>points of intersection</u> of the two functions.

To find the x-values of the points of intersection, equate the two functions and solve for x:

$\implies g(x)=f(x)$

$\implies x^2-9=-x+3$

$\implies x^2+x-12=0$

$\implies x^2+4x-3x-12=0$

$\implies x(x+4)-3(x+4)=0$

$\implies (x-3)(x+4)=0$

Apply the zero-product property:

$\implies (x-3)= \implies x=3$

$\implies (x+4)=0 \implies x=-4$

From inspection of the graph, we can see that the x-value of point D is <u>negative</u>, therefore the x-value of point D is x = -4.

To find the y-value of point D, substitute the found value of x into one of the functions:

$\implies f(-4)=-(-4)=7$

Therefore, D = (-4, 7).

The length of DE is the difference between the y-value of D and the x-axis:

⇒ DE = 7 units

The length of DF is the difference between the x-value of D and the x-axis:

⇒ DF = 4 units

To find point S, substitute the x-value of point T into function g(x):

$\implies g(4)=(4)^2-9=7$

Therefore, S = (4, 7).

The length ST is the difference between the y-values of points S and T:

$\implies ST=y_S-y_T=7-(-1)=8$

Therefore, ST = 8 units.

The given length of QR (⁴⁵/₄) is the difference between the functions at the same value of x. To find the x-value of points Q and R (and therefore the x-value of point M), subtract g(x) from f(x) and equate to QR, then solve for x:

$\implies f(x)-g(x)=QR$

$\implies -x+3-(x^2-9)=\dfrac{45}{4}$

$\implies -x+3-x^2+9=\dfrac{45}{4}$

$\implies -x^2-x+\dfrac{3}{4}=0$

$\implies -4\left(-x^2-x+\dfrac{3}{4}\right)=-4(0)$

$\implies 4x^2+4x-3=0$

$\implies 4x^2+6x-2x-3=0$

$\implies 2x(2x+3)-1(2x+3)=0$

$\implies (2x-1)(2x+3)=0$

Apply the zero-product property:

$\implies (2x-1)=0 \implies x=\dfrac{1}{2}$

$\implies (2x+3)=0 \implies x=-\dfrac{3}{2}$

As the x-value of points M, Q and P is negative, x = -³/₂.

Length OM is the difference between the x-values of points M and the origin O:

$\implies x_O-x_m=o-(-\frac{3}{2})=\dfrac{3}{2}$

Therefore, OM = ³/₂ units.

The values of x for which g(x) ≥ 0 are the values of x when the parabola is above the x-axis.

Therefore, g(x) ≥ 0 when x ≤ -3 and x ≥ 3.

8 0

11 months ago

Read 2 more answers