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3 years ago

PLEASE HELP!!!

Mathematics

2 answers:

goldfiish [28.3K]3 years ago

6 0

The correct answer to the first question is: A'(-5, 6)

The second answer is un-answerable, because the pictures you have provided do not display triangles.

Viktor [21]3 years ago

4 0

Answer:

For #3: The correct answer is A For #4: The correct answer is B

Step-by-step explanation:

A'(-5, 6)

Counterclockwise 90

Hope this helps! :)

Plz mark Brainliest trying to rank up!

Have a nice day!

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Select the two descriptions that match the dilation

postnew [5]

C and D

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3 years ago

using the midpoint formula find the coordinates of the midpoint 9f each sigment with end points (4,2),(-5,-1)

exis [7]

Answer:

The midpoint of the segment is (-0.5,0.5).

Step-by-step explanation:

Midpoint of a segment:

The midpoint of a segment is given by the mean of its coordinates.

Segment with end points (4,2),(-5,-1)

x-coordinates: 4 and -5

y-coordinates: 2 and -1

x-coordinate of the midpoint:

$x = \frac{4-5}{2} = -\frac{1}{2} = -0.5$

y-coordinate of the midpoint:

$y = \frac{2-1}{2} = \frac{1}{2} = 0.5$

The midpoint of the segment is (-0.5,0.5).

3 0

3 years ago

Given g(x)=4-2(3-x), what is the value of g(1)

avanturin [10]

Answer:

g(1) = 0

Step-by-step explanation:

To evaluate g(1) , substitute x = 1 into g(x), that is

g(1) = 4 - 2(3 - 1)

= 4 - 2(2)

= 4 - 4

= 0

3 0

3 years ago

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I own a large truck, and my neighbor owns four small trucks that are all identical. My truck can carry a load of at least $600$

erastova [34]

Answer:

there is no greatest load

Step-by-step explanation:

Let x and y represent the load capacities of my truck and my neighbor's truck, respectively. We are given two relations:

x ≥ y +600 . . . . . my truck can carry at least 600 pounds more

x ≤ (1/3)(4y) . . . . . my truck carries no more than all 4 of hers

Combining these two inequalities, we have ...

4/3y ≥ x ≥ y +600

1/3y ≥ 600 . . . . . . . subtract y

y ≥ 1800 . . . . . . . . multiply by 3

My truck's capacity is greater than 1800 +600 = 2400 pounds. This is a lower limit. The question asks for an <em>upper limit</em>. The given conditions do not place any upper limit on truck capacity.

4 0

3 years ago

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There are 3 red marbles, 5 green marbles, and 2 blue marbles in a jar. Event A is defined as drawing a red marble from the jar o

Anastaziya [24]

P(B/A) is $\frac{5}{9}$ in the simplest form

Step-by-step explanation:

Let us explain how to solve the problem

P(B|A) is called the "Conditional Probability" of B given A
Conditional probability is the probability of one event occurring with some relationship to one or more other events that means event A has already happened, now what is the chance of event B
The formula for conditional probability is P(B|A) = P(A and B)/P(A), where P(A and B) is P(A) . P(B given A occurred)

∵ There are 3 red marbles, 5 green marbles, and 2 blue

marbles in the jar

∴ The total number of the marbles in the jar = 3 + 5 + 2 = 10

∵ Two marbles are drawn from the bag, one after the other without

replacement

- That means red is already occurred when green is happened,

so it is a "Conditional Probability" of green given red

P(B/A) = P(A and B)/P(A)

∵ Event A is defined as drawing a red marble from the jar on

the first draw

∴ P(A) = P(red)

∵ P(red) = $\frac{3}{10}$

∴ P(A) = $\frac{3}{10}$

∵ Event B is defined as drawing a green marble on the second draw

∴ P(B) = P(green)

- The number of marbles after the red is chosen is 9

∵ P(green) = $\frac{5}{9}$

∴ P(B) = $\frac{5}{9}$

- Find P( A and B)

∴ P(A and B) = ( $\frac{3}{10}$ ) . ( $\frac{5}{9}$ )

∴ P(A and B) = $\frac{15}{90}$

- Simplify the fraction by divide up and down by 15

∴ P(A and B) = $\frac{1}{6}$

∵ P(B|A) = P(A and B)/P(A)

- Substitute the values of P(A and B) and P(A) in the rule above

∴ P(B/A) = $\frac{\frac{1}{6}}{\frac{3}{10}}$

- Simplify the fraction

∵ $\frac{1}{6}$ ÷ $\frac{3}{10}$ = $\frac{1}{6}$ × $\frac{10}{3}$ = $\frac{10}{18}$

- Divide up and down by 2

∴ P(B/A) = $\frac{5}{9}$

P(B/A) is $\frac{5}{9}$ in the simplest form

Learn more:

You can learn more about probability in brainly.com/question/10744457

#LearnwithBrainly

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3 years ago

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