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damaskus [11]
2 years ago
6

Need some help here IMMEDIATELY!! :(

Mathematics
1 answer:
ExtremeBDS [4]2 years ago
7 0

Answer: C option

Step-by-step explanation:

Since the drawing is 5x more than real object then we have to divide the numbers

a= 35/5 = 7

b= 30/5 = 6

c= 15/3 = 3

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If the width of a rectangle is x + 1 x and its length is 3x x2 - 1 , then its area is
Ronch [10]

Answer:

Area = 3x^5+3x^4-3x^3-3x^2

Step-by-step explanation:

Area equals length times width

x(x+1) * 3x(x^2-1)

(x^2+x) * (3x^3-3x)

3x^2(x+1)(x^2-1)

3x^5+3x^4-3x^3-3x^2

5 0
3 years ago
Read 2 more answers
find h and k such that the equation 3x^2 -hx +4k =0, so that the sum of the solutions is -12 and the product of the solutions is
Radda [10]
Let the solutions be a and b.
a = -2; b = -10
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(x + 2)(x + 10) = 0

x^2 + 10x + 2x + 20 = 0

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-h = 12

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6 0
3 years ago
The side length of aDavid is a statistician. He has a sample size of 40 (which he cannot change). What element of his hypothesis
Sergeeva-Olga [200]

Answer: D) the significance level of the test

=======================================================

Explanation:

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Going in the opposite direction, increasing the alpha value will make it more likely to reject the null. Each time you adjust the alpha value, keep the p-value to some fixed number (between 0 and 1).

6 0
3 years ago
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The flag of a country contains an isosceles triangle.​ (Recall that an isosceles triangle contains two angles with the same​ mea
Mrac [35]
Let us start with the unknown. Let us give one of the base angles the value of x. The other base angle is also x since both the base angles in an isosceles triangle will be equal. The remaining angle is 40 more than three times one of the base angles.It is 3x + 40. Here we get the equation

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5 0
4 years ago
Can anyone help me solve a trigonomic identity problem and also help me how to do it step by step?
dusya [7]
\bf cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}
\qquad csc(\theta)=\cfrac{1}{sin(\theta)}
\\\\\\
sin^2(\theta)+cos^2(\theta)=1\\\\
-------------------------------\\\\

\bf \cfrac{cos(\theta )cot(\theta )}{1-sin(\theta )}-1=csc(\theta )\\\\
-------------------------------\\\\
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\cfrac{cos^2(\theta )}{sin(\theta )}\cdot \cfrac{1}{1-sin(\theta )}-1
\\\\\\
\cfrac{cos^2(\theta )}{sin(\theta )[1-sin(\theta )]}-1\implies 
\cfrac{cos^2(\theta )-1[sin(\theta )[1-sin(\theta )]]}{sin(\theta )[1-sin(\theta )]}

\bf \cfrac{cos^2(\theta )-1[sin(\theta )-sin^2(\theta )]}{sin(\theta )[1-sin(\theta )]}\implies \cfrac{cos^2(\theta )-sin(\theta )+sin^2(\theta )}{sin(\theta )[1-sin(\theta )]}
\\\\\\
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3 years ago
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