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lawyer [7]
3 years ago
12

Find the length of each side of an equilateral triangle with perimeter 60 inches.

Mathematics
1 answer:
slava [35]3 years ago
8 0

Answer:

20

Step-by-step explanation:

60 divided into 3 = 20 since all sides are equal then you get 20 for each side.

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Anyone wanna help me? Ill mark brainliest
sdas [7]

Answer:

(a+11)(a-1)

Step-by-step explanation:

3 0
3 years ago
Help please due today
Vadim26 [7]

Answer:

Line a. goes to table 3, line b. goes to table 2, and line c. goes to table 1.

Step-by-step explanation:

Here are the 3 lines graphed (I even labeled each for you) so you can have a bit of a visual.

Hopefully you can find the points on each graph.

(Hint: The x row represents the x coordinate of an ordered pair, and the y row represents the y coordinate of an ordered pair.)

Ordered pairs look like this btw (x,y)

Hope this helps :)

6 0
2 years ago
Find the sum of each series, if it exists<br>91 + 85 + 79 + … + (­29)
Natali [406]

Answer:

651.

Step-by-step explanation:

Note: In the given series it should be -29 instead of 29 because 29 cannot be a term of AP whose first term is 91 and common difference is -6.

Consider the given series is

91+85+79+...+(-29)

It is the sum of an AP. Here,

First term = 91

Common difference = 85 - 91 = -6

Last term = -29

nth term of an AP is

a_n=a+(n-1)d

where, a is first term and d is common difference.

-29=91+(n-1)(-6)

-29-91=(n-1)(-6)

\dfrac{-120}{-6}=(n-1)

20=(n-1)

n=20+1=21

Sum of AP is

Sum=\dfrac{n}{2}[\text{First term + Last term}]

Sum=\dfrac{21}{2}[91+(-29)]

Sum=\dfrac{21}{2}[62]

Sum=651

Therefore, the sum of given series is 651.

5 0
2 years ago
Help me pleaseee.....
Lana71 [14]

Answer:

titutex=cos\alp,\alp∈[0:;π]

\displaystyle Then\; |x+\sqrt{1-x^2}|=\sqrt{2}(2x^2-1)\Leftright |cos\alp +sin\alp |=\sqrt{2}(2cos^2\alp -1)Then∣x+

1−x

2

∣=

2

(2x

2

−1)\Leftright∣cos\alp+sin\alp∣=

2

(2cos

2

\alp−1)

\displaystyle |\N {\sqrt{2}}cos(\alp-\frac{\pi}{4})|=\N {\sqrt{2}}cos(2\alp )\Right \alp\in[0\: ;\: \frac{\pi}{4}]\cup [\frac{3\pi}{4}\: ;\: \pi]∣N

2

cos(\alp−

4

π

)∣=N

2

cos(2\alp)\Right\alp∈[0;

4

π

]∪[

4

3π

;π]

1) \displaystyle \alp \in [0\: ;\: \frac{\pi}{4}]\alp∈[0;

4

π

]

\displaystyle cos(\alp -\frac{\pi}{4})=cos(2\alp )\dotscos(\alp−

4

π

)=cos(2\alp)…

2. \displaystyle \alp\in [\frac{3\pi}{4}\: ;\: \pi]\alp∈[

4

3π

;π]

\displaystyle -cos(\alp -\frac{\pi}{4})=cos(2\alp )\dots−cos(\alp−

4

π

)=cos(2\alp)…

1

Top

Display

6 0
3 years ago
Can someone please help me with this??
poizon [28]
Neither Will nor Olivia.
7 0
3 years ago
Read 2 more answers
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