All categories

3 years ago

1 Point

Mathematics

1 answer:

malfutka [58]3 years ago

7 0

Answer:

The solution of the system is

x=8, y=11

Step-by-step explanation:

we have

$y = 2x-5$ ----> equation A

$y = x+3$ ----> 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 point is (8,11)

see the attached figure

therefore

The solution of the system is

x=8, y=11

You might be interested in

Graph triangle XYZ with vertices X(0, -4), Y (4, -1), and Z(6, -3) on the coordinate grid.

gogolik [260]

Answer:

Step-by-step explanation:

6 0

3 years ago

Help me please it hardddd

iogann1982 [59]

There are a total 40 cars

and there is x yellow cars

and there are nine times as much white cars (9x white cars) lol

so x+9x=40

simplify 10x=40

divide both sides by ten 10x/10=40/10

x=4

we are looking for the number of WHITE cars sooo we multiply by 9-

9x=36

how is this "hardddd" no offence*

sorry im a mean person

3 0

3 years ago

Move two choices to the blanks to correctly complete the sentences.

Pavel [41]

Answer:

(A) There should have been 5 outcomes of HT

(B) The experimental probability is greater than the theoretical probability of HT.

Step-by-step explanation:

Given

$S = \{HH,HT,TH,TT\}$ -- Sample Space

$n(S) = 4$ --- Sample Size

Solving (a); theoretical outcome of HT in 20 tosses

First, calculate the theoretical probability of HT

$P(HT) = \frac{n(HT)}{n(S)}$

$P(HT) = \frac{1}{4}$

Multiply this by the number of tosses

$P(HT) * n= \frac{1}{4} * 20$

$P(HT) * n= 5$

Solving (b); experimental probability of HT

Here, we make use of the table

$P(HT) = \frac{n(HT)}{n(S)}$

$P(HT) = \frac{6}{20}$

$P(HT) = 0.30$ ---- Experimental Probability

In (a), the theoretical probability is:

$P(HT) = \frac{1}{4}$

$P(HT) = 0.25$ ---- Experimental Probability

By comparison;

$0.30 > 0.25$

6 0

3 years ago

Enter the explicit rule for the geometric sequence.<br><br> 9,6,4,83,…

soldier1979 [14.2K]

Answer:

a_n = 2^(n - 1) 3^(3 - n)

Step-by-step explanation:

9,6,4,8/3,…

a1 = 3^2

a2 = 3 * 2

a3 = 2^2

As we can see, the 3 ^x is decreasing and the 2^ y is increasing

We need to play with the exponent in terms of n

Lets look at the exponent for the base of 2

a1 = 3^2 2^0

a2 = 3^1 2^1

a3 = 3^ 0 2^2

an = 3^ 2^(n-1)

I picked n-1 because that is where it starts 0

n = 1 (1-1) =0

n=2 (2-1) =1

n=3 (3-1) =2

Now we need to figure out the exponent for the 3 base

I will pick (3-n)

n =1 (3-1) =2

n =2 (3-2) =1

n=3 (3-3) =0

8 0

3 years ago

Circle sector type question with figure.

igomit [66]

Can’t help if there’s no figure

4 0

3 years ago