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lozanna [386]
3 years ago
8

Solve photo question. Mathematics. Please.

Mathematics
1 answer:
guajiro [1.7K]3 years ago
8 0
The second dot with says x+y=360 y=8x
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HELP ASAP/ WILL MARK BRAINLYIEST If f(x) = 3x + 2, what is f(5)?
krok68 [10]
F(5) is equal to 17.
f(5)= 3(5)+2= 17
4 0
3 years ago
Use the figure above to answer the questions.
DENIUS [597]

Answer:

the answer is d

Step-by-step explanation:

the shape of it as well

4 0
3 years ago
John often speed while driving to school in order to arrive on time. The probability that he will speed to school is 0.75. If th
BartSMP [9]

Answer:

0.33

Step-by-step explanation:

Given the following :

P(speeding) = p(s) = 0.75

P(being stopped) = p(t)

P(speeding and gets stopped) = p(s n t) = 0.25

Find the probability that he is stopped, given that he is speeding is written as P(t | s) ;

P(t | s) = p(s n t) / p(s)

P(s n t) = 0.25

P(s) = 0.75

Hence,

P(t | s) = 0.25 / 0.75

P(t | s) = 0.33

3 0
3 years ago
Ashleys Shoe Store buys shoes for $35 and then sells them for
Gelneren [198K]
85/35 = 2.43% because that is the percent you would need to multiple by
3 0
2 years ago
Read 2 more answers
Which side lengths form a right triangle? Choose all answers that apply: Choose all answers that apply: (Choice A) A \sqrt4, 10,
Vika [28.1K]

Answer:

(C)      

Step-by-step explanation:

In order to find whether the given side lengths form a right triangle or not, we must show that it satisfies the Pythagoras theorem.

A. \sqrt{4}, 10 and 116

According to Pythagoras theorem,

⇒(116)^2=4+100

⇒13456=104

which does not satisfy the condition, therefore these side lengths do not form a right triangle.

B. 3,6 and 8

According to Pythagoras theorem,

⇒(8)^{2}=(3)^2+(6)^2

⇒64=9+36

⇒64=45

which does not satisfy the condition, therefore these side lengths do not form a right triangle.

C. 30, 40 and 50

According to Pythagoras theorem,

⇒(50)^{2}=(30)^2+(40)^2

⇒2500=900+1600

⇒2500=2500

which satisfies the condition, therefore theses side lengths form right triangle.

5 0
3 years ago
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