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morpeh [17]
3 years ago
8

The formula d=rt relates three quantities distance (d) rate (r) and time (t)

Mathematics
2 answers:
algol [13]3 years ago
8 0

Here you go, this should be right.

polet [3.4K]3 years ago
7 0
B)5

c) aprox. 55.5 or 55.6
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Am I correct? Will give brainliest
Korolek [52]

twice the sum of a number and 3, means you need to add the number to 3 first, then multiply by 2.

The correct expression would be 2(n+3)

4 0
3 years ago
Which two ratios represent quantities that are proportional?
Ad libitum [116K]

Answer:

All 4 of them

Step-by-step explanation:

Tell me if I'm wrong

3 0
2 years ago
A number is equal to twice a smaller number plus 3. Yhe same number is equal to twice the sum of smaller number and 1. How many
allsm [11]

Answer:

There is no solution to the problem

Step-by-step explanation:

In order to find the solutions to the problem you can write the situation in an algebraic form.

You have that a number is equal to twice a smaller number plus 3. This can be written as follow:

x=2y+3      (1)

where x is the number and y is the smaller number

Furthermore, the same number x is equal to twice the sum of the smaller number and 1, which can be written as follow:

x=2(y+1)     (2)

To find the solution you equal the equation (1) with (2), and you solve for y:

2y+3=2y+1

3=1

You obtain an inconsistency, hence, the situation of the problem does not have solution

5 0
3 years ago
(x+2)(x-3)(x+5) is identical to x^(3)+ax^(2)-11x+b<br> Find the value of a and the value of b
lutik1710 [3]

Answer:

a = 4 and b = - 30

Step-by-step explanation:

Expand the left side and compare like terms on both sides, that is

(x + 2)(x - 3)(x + 5) ← expand the first pair of factors using FOIL

= (x² - x - 6)(x + 5) ← distribute

= x³ + 5x² - x² - 5x - 6x - 30 ← collect like terms

= x³ + 4x² - 11x - 30

Compare like terms with x³ +ax² - 11x + b

4x² and ax² ⇒ a = 4

+ b and - 30 ⇒ b = - 30

3 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
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