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Gemiola [76]
2 years ago
12

PLEASE HELP FAST I NEED THISSSS

Mathematics
2 answers:
masya89 [10]2 years ago
6 0

Answer:

10.26

im bad at explaining

agasfer [191]2 years ago
3 0
Area of a square is length squared and of a triangle is (length x width)/2. Also keep in mind we can make it 8 triangles with the length and width as 6 and 0.4ft so...
0.8 ^2 = 0.64
6 x 0.4 = 2.4 x 8 =19.2
19.2 + 0.64 = 19.84
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Kyle rides his bicycle 15 mph for 2 hours how far does he travel
Cerrena [4.2K]

Answer:

30mph

Step-by-step explanation:

Kyle rides his bike 30mph,

15mph x 2hrs = 30 mph.

15 miles per hour x 3 is 30mph total.

Hope this helps!

4 0
3 years ago
Read 2 more answers
Which of the following values of x are solutions to the equation 4x^2+x-60=0​
Vladimir79 [104]

Answer:

x = 15/4, - 4

Step-by-step explanation:

I'm not sure :p

5 0
2 years ago
PLZZZZ HELPPPPPPPPP!!!! a survey asks teachers and students whether they would like the new school mascot to be a Viking or a Pa
VARVARA [1.3K]

Answer:

B

Step-by-step explanation:

Two events A and B are called independent, when

P(B|A)=P(B),

where P(B|A) is the probability that B occurs given that A has already occurred.

A = randomly selected person is a student

B = randomly selected person prefer "Vikings"

B|A = randomly selected student prefer "Vikings"

Probabilities:

P(A)=\dfrac{100}{125}=\dfrac{4}{5}=0.8\\ \\P(B|A)=\dfrac{80}{85}=\dfrac{16}{17}\approx 0.94

Since

0.80\neq 0.94

we can state that events are not independent.

8 0
3 years ago
Read 2 more answers
8x+3y=18
Minchanka [31]

Step-by-step explanation:

8x+3y=18

subject y according y=mx+c

y=8/3x-6

(-9,-13)

substitute -9 to y=8/3x-6

-13=8/3×(-9)+c

11=c

y=8/3x+11

3 0
3 years ago
The quotient of negative eight and the sum of a number and three
PolarNik [594]
For this case, the first thing we must do is define a variable.
 We have then:
 x: real number
 Now we write the expression:
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 The number x can be all real different from minus three.
 Answer:
 
-8 / (x + 3)
 
With x different from -3.
6 0
3 years ago
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