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Inessa [10]
3 years ago
11

(2b^2 - 5b) - (7b + 3b^2)

Mathematics
1 answer:
Marta_Voda [28]3 years ago
8 0

Answer:

-13b

Step-by-step explanation:

(2b^2 - 5b) - (7b + 3b^2)

-3b^2 - 10b^2 = -13b

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Are similar polygons congruent?
Vilka [71]

Polygons are similar when they have the same number of sides,
their corresponding angles are equal, and their corresponding
sides are all in the same ratio.

When that ratio is ' 1 ', they are also congruent.

That can happen sometimes, but it doesn't always have to happen.

5 0
3 years ago
Read 2 more answers
option 1 drop down are: even-odd identity, quotient identity, Pythagorean identity, double-number identity.option 2 drop down ar
motikmotik

Answer:

The equation is given below as

\frac{\cos2x}{\cos x}=\cos x-\sin x\tan x

Step 1:

We will work on the left-hand side, we will have

\begin{gathered} \cos x-\sin x\tan x \\ \text{recall that,} \\ Quoitent\text{ identity is} \\ \tan x=\frac{\sin x}{\cos x} \end{gathered}

By substituting the identity above, we will have

\begin{gathered} \cos x-\sin x\tan x=\cos x-\frac{\sin x.\sin x}{\cos x}=\cos x-\frac{\sin^2x}{\cos x} \\  \end{gathered}

Here, we will make use of the quotient identity

Step 2:

By writings an expression, we will have

\begin{gathered} \cos x-\sin x\tan x=\cos x-\frac{\sin x.\sin x}{\cos x} \\ \cos x-\sin x\tan x=\frac{\cos^2x-\sin^2x}{\cos x} \end{gathered}

Here, we will use the definition of subtraction

\cos x-\frac{\sin^2x}{\cos x}

Step 3:

We will apply the double number identity given below

\begin{gathered} \cos 2\theta=\cos (\theta+\theta)=\cos ^2\theta-\sin ^2\theta \\ \cos 2x=cos(x+x)=\cos ^2x-\sin ^2x \end{gathered}

By applying this, we will have

\frac{\cos^2x-\sin^2x}{\cos x}=\frac{\cos2x}{\cos x}

Here, we will use the double number identity

\frac{\cos^2x-\sin^2x}{\cos x}

5 0
1 year ago
Kendra hiked each day for a week the first day she hiked one over at 8 mile the second day she hiked three of an 8 mile on the t
Reil [10]
I'm not sure I'm not in that level yet
6 0
3 years ago
Find the sum of the positive integers less than 200 which are not multiples of 4 and 7​
taurus [48]

Answer:

12942 is the sum of positive integers between 1 (inclusive) and 199 (inclusive) that are not multiples of 4 and not multiples 7.

Step-by-step explanation:

For an arithmetic series with:

  • a_1 as the first term,
  • a_n as the last term, and
  • d as the common difference,

there would be \displaystyle \left(\frac{a_n - a_1}{d} + 1\right) terms, where as the sum would be \displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n).

Positive integers between 1 (inclusive) and 199 (inclusive) include:

1,\, 2,\, \dots,\, 199.

The common difference of this arithmetic series is 1. There would be (199 - 1) + 1 = 199 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}.

Similarly, positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 4 include:

4,\, 8,\, \dots,\, 196.

The common difference of this arithmetic series is 4. There would be (196 - 4) / 4 + 1 = 49 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 7 include:

7,\, 14,\, \dots,\, 196.

The common difference of this arithmetic series is 7. There would be (196 - 7) / 7 + 1 = 28 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 28 (integers that are both multiples of 4 and multiples of 7) include:

28,\, 56,\, \dots,\, 196.

The common difference of this arithmetic series is 28. There would be (196 - 28) / 28 + 1 = 7 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}.

The requested sum will be equal to:

  • the sum of all integers from 1 to 199,
  • minus the sum of all integer multiples of 4 between 1\! and 199\!, and the sum integer multiples of 7 between 1 and 199,
  • plus the sum of all integer multiples of 28 between 1 and 199- these numbers were subtracted twice in the previous step and should be added back to the sum once.

That is:

19900 - 4900 - 2842 + 784 = 12942.

8 0
3 years ago
Three students need to produce a prime factorization of 48. Donna states that the first factors in the tree should be 6 and 8. L
Gennadij [26K]
If we start with 6 and 8, we can break 6 up into 2*3 and 8 into 2*2*2, thus getting a prime factorization of 2*2*2*2*3, or 2^4 *3.

If we begin with 4 and 12, 4 breaks into 2*2 and 12 into 2*2*3, so the prime factorization of 48 is still 2^4 *3.

The starting factors do not matter, since the answer comes out to be the same. There is exactly one correct answer- it doesn't matter how it's found.

Hope this helps! :)
4 0
4 years ago
Read 2 more answers
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