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3 years ago

Can someone give me the workings out for this one please

Mathematics

1 answer:

Elza [17]3 years ago

5 0

50mph, 60 minutes, half both 25 and 30
A-B should be 30 minutes
B-C should probably be 20-25 personally I'd go with 25
so 55?

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in an ap,the first term of an arithmetic series is 5,the last term is 45 and the sum is 400. find the number of terms and common

Olegator [25]

Answer:

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3 years ago

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Please help me idk this

tatiyna

Answer:

2.25

Step-by-step explanation:

The question gives us one hint. That side of the square is 1.5"

Because it gives us that, we can use that to find the area. Because it is a square, all of the sides are the same length.

To find the area it is basexheight

So that is why the answer is 2.25 because 1.5x1.5=the area of the square

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Which rule applies to the translation of the RED trapezoid to the BLUE trapezoid? A) (x, y) → (x + 2, y − 3) B) (x, y) → (x + 4,

kicyunya [14]

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The perimeter of a rectangle is 152m. If the length is one morethan twice of its width. Find the length and width of rectangle.

sergejj [24]

Step-by-step explanation:

Perimeter of a Rect.= 2(l+b)

b= x

l= 2x

Equation

152= 2(2x+x)

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3 years ago

How do you this question (pre-cal)

Novay_Z [31]

The goal to proving identities is to transform one side into the other. We can only pick one side to transform while the other side stays the same the entire time. The general rule of thumb is to transform the more complicated side (though there may be exceptions to this guideline).

So I'll take the left hand side and try to turn it into $\csc^2( B )$

One way we can do that is through the following steps:

$\frac{\tan(B) + \cot(B)}{\tan(B)} = \csc^2(B)\\\\\frac{\tan(B)}{\tan(B)} + \frac{\cot(B)}{\tan(B)} = \csc^2(B)\\\\1 + \cot(B)*\frac{1}{\tan(B)} = \csc^2(B)\\\\1 + \cot(B)*\cot(B) = \csc^2(B)\\\\1 + \cot^2(B) = \csc^2(B)\\\\1 + \frac{cos^2(B)}{\sin^2(B)} = \csc^2(B)\\\\\frac{sin^2(B)}{\sin^2(B)}+\frac{cos^2(B)}{\sin^2(B)} = \csc^2(B)\\\\\frac{sin^2(B)+cos^2(B)}{\sin^2(B)} = \csc^2(B)\\\\\frac{1}{\sin^2(B)} = \csc^2(B)\\\\\csc^2(B)=\csc^2(B) \ \ {\Large \checkmark}\\\\$

Since we've shown that the left hand side transforms into the right hand side, this verifies the equation is an identity.

4 0

3 years ago