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Shalnov [3]
3 years ago
14

The graphs of y=x2- 3 and y = 3x -4 intersect at

Mathematics
1 answer:
OLga [1]3 years ago
8 0

Answer:

Option 3) (0.38, -2.85) and (2.62,3.85)

Step-by-step explanation:

we have

y=x^{2}-3 ----> equation A

y=3x-4 ----> equation B

Solve the system by graphing

Remember that the solution of the system of equations is the intersection point both graphs

using a graphing tool

The intersection points are (0.38,-2.85) and (2.62,3.85)

see the attached figure

therefore

The graphs intersect approximately at (0.38,-2.85) and (2.62,3.85)

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Evaluate the expression below if m = 350.<br><br> 3.5m =
Lena [83]

Answer:

1,225

Step-by-step explanation:

3.5×350=1225

its saying m is 350 and the question is asking 3.5 times 350

8 0
3 years ago
Please help.
serious [3.7K]

The number of throws that the player make X follows binomial distribution.

The probability that the player makes x throws out of eleven throws is

P(X=x)=C(x,11)0.5^x(0.5)^{11-x}=C(x,11)0.5^{11}

The probability that the player makes at most nine out of eleven free throws

is

\sum_{x=0}^9C(x,11)0.5^{11}=1-\sum_{x=10}^{11}C(x,11)0.5^{11}\\ =1-C(10,11)0.5^{11}-C(11,11)0.5^{11}=1-\frac{12}{2^{11}} \\ =1-\frac{3}{512} =\frac{509}{512}

Correct choice is (B).

8 0
3 years ago
The owner of the Rancho Grande has 3,044 yd of fencing with which to enclose a rectangular piece of grazing land situated along
inysia [295]

Answer:

Step-by-step explanation:

Area of the rectangular piece A = xy

x is the length

y is the width

Perimeter =2x+2y

Given

Perimeter = 3044yds

3044 = 2x+2y

1522 = x+y

x = 1522-y

Substitute into the formula for area

A = xy

A = (1522-y)y

A = 1522y-y²

To maximize the area, dA/dy must be zero

dA/dy = 1522-2y

0 = 1522-2y

2y = 1522

y = 1522/2

y = 761

Since x+y = 1522

X+761 = 1522

X = 761

Dimension is 761yd by 761yd

Area = 761×761 = 579121yd²

7 0
3 years ago
Hi guys, can anyone help me with this triple integral? Many thanks:)
Crank

Another triple integral.  We're integrating over the interior of the sphere

x^2+y^2+z^2=2^2

Let's do the outer integral over z.   z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

y^2=4-x^2-z^2

x is the inner integral so at this point we conservatively say its zero.  That means y goes from -\sqrt{4-z^2} and +\sqrt{4-z^2}

Similarly the inner integral x goes between \pm-\sqrt{4-y^2-z^2}

So we rewrite the integral

\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz

Let's work on the inner one,

\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz

There's no z in the integrand, so we treat it as a constant.

=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}

So the middle integral is

\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy  

I gotta go so I'll stop here, sorry.

7 0
3 years ago
Read 2 more answers
PLEASE HELP WITH ALL!!!
rosijanka [135]
3/4. triangle FHE and triangle KMN; triangle GEH and triangle LNM
3 0
3 years ago
Read 2 more answers
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