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3 years ago

Thank you very much for your help!

Mathematics

1 answer:

Kobotan [32]3 years ago

5 0

For a.
$x^2+y^2=25$

For b.
$(x-4)^2+(y-3)^2=25$

For c.
$(x-5)^2+(y-4)^2=9$

Hope that helps

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Can you help on word problems?

Ne4ueva [31]

Uh sure if you give me the problem I could help

4 0

3 years ago

The length of a radius of a circle, measured in centimeters, is represented by the expression x + 1.5. The diameter of the circl

Tomtit [17]

The value of x of a circle with radius x + 1.5 and diameter of 9 2 / 5 cm is 7.9 cm

<h3>Radius of a circle</h3>

Radius of a circle is half of the diameter of a circle. The radius extend from the centre of the circle to the circumference.

Mathematically,

radius = 1 / 2 (diameter)

Therefore,

radius = (x + 1.5) cm

diameter = 9 2 / 5 cm = 47 / 5 cm

let's find x with the relationship above.

Therefore,

x + 1.5 = 47 / 5

subtract 1.5 from both sides

x + 1.5 - 1.5 = 9.4 - 1.5

x = 7.9 cm

learn more on radius here: brainly.com/question/12183018

6 0

2 years ago

All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be

Aneli [31]

Answer:

There is a 2% probability that the student is proficient in neither reading nor mathematics.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a student is proficient in reading

B is the probability that a student is proficient in mathematics.

C is the probability that a student is proficient in neither reading nor mathematics.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a student is proficient in reading but not mathematics and $A \cap B$ is the probability that a student is proficient in both reading and mathematics.

By the same logic, we have that:

$B = b + (A \cap B)$

Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So

$(A \cup B) + C = 1$