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3 years ago

What would the result be if a was increased by 25%

Mathematics

2 answers:

strojnjashka [21]3 years ago

6 0

Answer:

you forgot to add a picture

bezimeni [28]3 years ago

3 0

Answer:

I believe you forgot a picture or something by the same lines

Step-by-step explanation:

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Which is a graph of the equation y = 2x - 4

AURORKA [14]

<h2>The third graph</h2><h3 /><h3>The graph has a slope of 2</h3><h3>and a y-intercept of -4</h3>

6 0

3 years ago

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What is the range of f(x) = – 4?

Andre45 [30]

The answer is -4. The range is the y-value and when you say f(x) that is the same thing as y, so if f(x) = -4 that means y = -4 which means the range is -4.

5 0

4 years ago

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A table is in the shape of a regular pentagon with a side length of 3 feet and an

ehidna [41]

Answer:

15.5 ft²

Step-by-step explanation:

The area (A) of a regular polygon is calculated as

A = $\frac{1}{2}$ pa ( p is the perimeter and a the apothem )

Here side of regular pentagon is 3 ft, thus

perimeter = 5 × 3 = 15, so

A = 0.5 × 15 × 2.06 ≈ 15.5 ft ( to the nearest tenth )

8 0

3 years ago

Simplify the expression. (3.5x - 5) + (7.2x + 10)

Andru [333]

Answer: 10.7x + 5

Step-by-step explanation:

(3.5 + 7.2)x + (-5 + 10) = 10.7x + 5

8 0

3 years ago

) f) 1 + cot²a = cosec²a

notsponge [240]

Answer:

It is an identity, proved below.

Step-by-step explanation:

I assume you want to prove the identity. There are several ways to prove the identity but here I will prove using one of method.

First, we have to know what cot and cosec are. They both are the reciprocal of sin (cosec) and tan (cot).

$\displaystyle \large{\cot x=\frac{1}{\tan x}}\\\displaystyle \large{\csc x=\frac{1}{\sin x}}$

csc is mostly written which is cosec, first we have to write in 1/tan and 1/sin form.

$\displaystyle \large{1+(\frac{1}{\tan x})^2=(\frac{1}{\sin x})^2}\\\displaystyle \large{1+\frac{1}{\tan^2x}=\frac{1}{\sin^2x}}$

Another identity is:

$\displaystyle \large{\tan x=\frac{\sin x}{\cos x}}$

Therefore:

$\displaystyle \large{1+\frac{1}{(\frac{\sin x}{\cos x})^2}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{1}{\frac{\sin^2x}{\cos^2x}}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}$

Now this is easier to prove because of same denominator, next step is to multiply 1 by sin^2x with denominator and numerator.

$\displaystyle \large{\frac{\sin^2x}{\sin^2x}+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}\\\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}$

Another identity:

$\displaystyle \large{\sin^2x+\cos^2x=1}$

Therefore:

$\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}\longrightarrow \boxed{ \frac{1}{\sin^2x}={\frac{1}{\sin^2x}}}$

Hence proved, this is proof by using identity helping to find the specific identity.

6 0

3 years ago