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Mumz [18]
4 years ago
14

Renee adds 5 to a number, then multiplies the sum by-2. The result is 8. Write and solve an equation to find the number, x. What

is the number? ​
Mathematics
1 answer:
miskamm [114]4 years ago
5 0
5+(-9)= -4 • -2 =8
i hope i’m right.
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Mademuasel [1]
The first option, as it positively supports the claim stated.
7 0
3 years ago
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Describe a situation that the expression -15÷(-15) can represent
FrozenT [24]
-15 so that is negative. so that is negative 15÷15 so 15 ÷15 is 1 so it would be -15÷-15=-1
7 0
3 years ago
Fill in the missing number in this newspaper report:
natka813 [3]

Answer:

£125000

Step-by-step explanation:

Original cost of house = £100 000

Percent increase = 25%

Increase in price = 25% of 100,000

Increase in price = 0.25 * 100,000

Increase in price = 25000

New cost = original cost + Increment

New cost = 100,000 + 25,000

New cost = 125,000

Hence it now cost  £125000

8 0
3 years ago
How to know if a function is periodic without graphing it ?
zhenek [66]
A function f(t) is periodic if there is some constant k such that f(t+k)=f(k) for all t in the domain of f(t). Then k is the "period" of f(t).

Example:

If f(x)=\sin x, then we have \sin(x+2\pi)=\sin x\cos2\pi+\cos x\sin2\pi=\sin x, and so \sin x is periodic with period 2\pi.

It gets a bit more complicated for a function like yours. We're looking for k such that

\pi\sin\left(\dfrac\pi2(t+k)\right)+1.8\cos\left(\dfrac{7\pi}5(t+k)\right)=\pi\sin\dfrac{\pi t}2+1.8\cos\dfrac{7\pi t}5

Expanding on the left, you have

\pi\sin\dfrac{\pi t}2\cos\dfrac{k\pi}2+\pi\cos\dfrac{\pi t}2\sin\dfrac{k\pi}2

and

1.8\cos\dfrac{7\pi t}5\cos\dfrac{7k\pi}5-1.8\sin\dfrac{7\pi t}5\sin\dfrac{7k\pi}5

It follows that the following must be satisfied:

\begin{cases}\cos\dfrac{k\pi}2=1\\\\\sin\dfrac{k\pi}2=0\\\\\cos\dfrac{7k\pi}5=1\\\\\sin\dfrac{7k\pi}5=0\end{cases}

The first two equations are satisfied whenever k\in\{0,\pm4,\pm8,\ldots\}, or more generally, when k=4n and n\in\mathbb Z (i.e. any multiple of 4).

The second two are satisfied whenever k\in\left\{0,\pm\dfrac{10}7,\pm\dfrac{20}7,\ldots\right\}, and more generally when k=\dfrac{10n}7 with n\in\mathbb Z (any multiple of 10/7).

It then follows that all four equations will be satisfied whenever the two sets above intersect. This happens when k is any common multiple of 4 and 10/7. The least positive one would be 20, which means the period for your function is 20.

Let's verify:

\sin\left(\dfrac\pi2(t+20)\right)=\sin\dfrac{\pi t}2\underbrace{\cos10\pi}_1+\cos\dfrac{\pi t}2\underbrace{\sin10\pi}_0=\sin\dfrac{\pi t}2

\cos\left(\dfrac{7\pi}5(t+20)\right)=\cos\dfrac{7\pi t}5\underbrace{\cos28\pi}_1-\sin\dfrac{7\pi t}5\underbrace{\sin28\pi}_0=\cos\dfrac{7\pi t}5

More generally, it can be shown that

f(t)=\displaystyle\sum_{i=1}^n(a_i\sin(b_it)+c_i\cos(d_it))

is periodic with period \mbox{lcm}(b_1,\ldots,b_n,d_1,\ldots,d_n).
4 0
3 years ago
The vertices of a triangle are A(0, 0), B(3, 8), and C(9, 0). What is the area of this triangle?
Harlamova29_29 [7]
Answer: 36 units squared.


Explanation:
Area of a traingle is given by,

| 0.5{x1(y2-y3) + x2(y3-y1) + x3(y1-y2)} |

Accepting the values of the coordinates in x's and y's,

x1 =0, y1=0

x2=3, y2=8

x3 =9, y3=0

= | 0.5 × (-72)|
= 36 unit squared.
8 0
4 years ago
Read 2 more answers
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