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-BARSIC- [3]
3 years ago
11

Which of the following number sentences is an example of the Distributive Property?

Mathematics
1 answer:
leonid [27]3 years ago
7 0

Answer:

d. 7 × (10 + 9) = (7 × 10) + (7 × 9)

Step-by-step explanation:

The format of writing numbers in distributive property is a(b+c) = (a×b)+(a×c)

We can see that in choice d. 7 × (10 + 9) = (7 × 10) + (7 × 9) it is in correct format of writing numbers in distributive property. Thus, choice d. is correct.

Verification:-

a(b+c) = (a×b)+(a×c)

7 × (10 + 9) = (7 × 10) + (7 × 9)

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Jeremy invested $60,000 earning 6.5% simple interest. How much interest will he earn over 5 years?
const2013 [10]

Answer:

$19500

Step-by-step explanation:

Interest = PRT/100

Where

P = principal

R = rate

T = time

Given

P = $60000

R = 6.5%

T = 5 yrs

Interest = 60000 x 6.5 x 5 /100

Multiply through

1950000/100

19500

$19500

He will earn $19500 interest over 5 years

7 0
4 years ago
In ΔSTU, s = 6.1 cm, t = 3.2 cm and u=7.4 cm. Find the measure of ∠S to the nearest 10th of a degree.
GuDViN [60]

Answer:

m∠S = 54.1°

Step-by-step explanation:

Use the law of cosines

s^{2}  = t^{2} + u^{2}  - 2utcos S\\6.1^{2} = 3.2^{2}  + 7.4^{2}  - 2(3.2)(7.4) cos S

37.21 = 10.24 + 54.76 - 47.36 cos S

-27.79 = -47.36 cos S

.586782 = cos S

arccos .586782 = S

m∠S = 54.1°

8 0
3 years ago
Read 2 more answers
Point R divides in the ratio 1 : 3. If the x-coordinate of R is -1 and the x-coordinate of P is -3, what is the x-coordinate of
Anon25 [30]

The x coordinate of Q would be 5.

In order to find this, first note that P and R are exactly 2 away from each other (-1 to -3 is two away).

Now, we know that Q is 3 times as far away from R as P is given the ratio at the beginning. Therefore we know it is 6 away from R (2 * 3 = 6).

Now we simply add 6 to the R value to get the Q value.

-1 + 6 = 5

3 0
3 years ago
For the function defined by f(t)=2-t, 0≤t<1, sketch 3 periods and find:
Oksi-84 [34.3K]
The half-range sine series is the expansion for f(t) with the assumption that f(t) is considered to be an odd function over its full range, -1. So for (a), you're essentially finding the full range expansion of the function

f(t)=\begin{cases}2-t&\text{for }0\le t

with period 2 so that f(t)=f(t+2n) for |t| and integers n.

Now, since f(t) is odd, there is no cosine series (you find the cosine series coefficients would vanish), leaving you with

f(t)=\displaystyle\sum_{n\ge1}b_n\sin\frac{n\pi t}L

where

b_n=\displaystyle\frac2L\int_0^Lf(t)\sin\frac{n\pi t}L\,\mathrm dt

In this case, L=1, so

b_n=\displaystyle2\int_0^1(2-t)\sin n\pi t\,\mathrm dt
b_n=\dfrac4{n\pi}-\dfrac{2\cos n\pi}{n\pi}-\dfrac{2\sin n\pi}{n^2\pi^2}
b_n=\dfrac{4-2(-1)^n}{n\pi}

The half-range sine series expansion for f(t) is then

f(t)\sim\displaystyle\sum_{n\ge1}\frac{4-2(-1)^n}{n\pi}\sin n\pi t

which can be further simplified by considering the even/odd cases of n, but there's no need for that here.

The half-range cosine series is computed similarly, this time assuming f(t) is even/symmetric across its full range. In other words, you are finding the full range series expansion for

f(t)=\begin{cases}2-t&\text{for }0\le t

Now the sine series expansion vanishes, leaving you with

f(t)\sim\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi t}L

where

a_n=\displaystyle\frac2L\int_0^Lf(t)\cos\frac{n\pi t}L\,\mathrm dt

for n\ge0. Again, L=1. You should find that

a_0=\displaystyle2\int_0^1(2-t)\,\mathrm dt=3

a_n=\displaystyle2\int_0^1(2-t)\cos n\pi t\,\mathrm dt
a_n=\dfrac2{n^2\pi^2}-\dfrac{2\cos n\pi}{n^2\pi^2}+\dfrac{2\sin n\pi}{n\pi}
a_n=\dfrac{2-2(-1)^n}{n^2\pi^2}

Here, splitting into even/odd cases actually reduces this further. Notice that when n is even, the expression above simplifies to

a_{n=2k}=\dfrac{2-2(-1)^{2k}}{(2k)^2\pi^2}=0

while for odd n, you have

a_{n=2k-1}=\dfrac{2-2(-1)^{2k-1}}{(2k-1)^2\pi^2}=\dfrac4{(2k-1)^2\pi^2}

So the half-range cosine series expansion would be

f(t)\sim\dfrac32+\displaystyle\sum_{n\ge1}a_n\cos n\pi t
f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}a_{2k-1}\cos(2k-1)\pi t
f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}\frac4{(2k-1)^2\pi^2}\cos(2k-1)\pi t

Attached are plots of the first few terms of each series overlaid onto plots of f(t). In the half-range sine series (right), I use n=10 terms, and in the half-range cosine series (left), I use k=2 or n=2(2)-1=3 terms. (It's a bit more difficult to distinguish f(t) from the latter because the cosine series converges so much faster.)

5 0
4 years ago
(b) Solve and write the solution set<br>for: 3x -8&lt;7 ; XEN<br>​
bekas [8.4K]

Answer:

x=5is greater than 7

hsiehdje

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6 0
3 years ago
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