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Salsk061 [2.6K]
3 years ago
12

Can some one please help me with this math assignment . I would be really grateful

Mathematics
2 answers:
Allushta [10]3 years ago
8 0

Answer: I need a picture

Step-by-step explanation:

RoseWind [281]3 years ago
5 0
Where is the assignment?
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The points plotted below are on the graph of a polynomial. Some of the roots to this polynomial are integers. Which of the follo
Shkiper50 [21]

Answer:

I think its B but I have very little evidence. Don't even put B but I was just saying I think it is.

Step-by-step explanation:

3 0
3 years ago
I am lost on what to do
Neko [114]
\bf sin({{ \alpha}})sin({{ \beta}})=\cfrac{1}{2}[cos({{ \alpha}}-{{ \beta}})\quad -\quad cos({{ \alpha}}+{{ \beta}})]
\\\\\\
cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}\\\\
-----------------------------\\\\
\lim\limits_{x\to 0}\ \cfrac{sin(11x)}{cot(5x)}\\\\
-----------------------------\\\\
\cfrac{sin(11x)}{\frac{cos(5x)}{sin(5x)}}\implies \cfrac{sin(11x)}{1}\cdot \cfrac{sin(5x)}{cos(5x)}\implies \cfrac{sin(11x)sin(5x)}{cos(5x)}

\bf \cfrac{\frac{cos(11x-5x)-cos(11x+5x)}{2}}{cos(5x)}\implies \cfrac{\frac{cos(6x)-cos(16x)}{2}}{cos(5x)}
\\\\\\
\cfrac{cos(6x)-cos(16x)}{2}\cdot \cfrac{1}{cos(5x)}\implies \cfrac{cos(6x)-cos(16x)}{2cos(5x)}
\\\\\\
\lim\limits_{x\to 0}\ \cfrac{cos(6x)-cos(16x)}{2cos(5x)}\implies \cfrac{1-1}{2\cdot 1}\implies \cfrac{0}{2}\implies 0
4 0
3 years ago
3 1/8t=2 1/2 what is t?​
navik [9.2K]

Answer:

T=4

Step-by-step explanation:

it is just half of what the other side is if one side is like 1 1/2 then you just break the other side in half and that will give you T.

5 0
3 years ago
Find the distance between the points 5 , 3 and -6 , 3 .
just olya [345]
(5, 3) (-6, 3)
The slope is 0 Slope = m = (Y2<span> -Y</span>1<span>) </span>÷<span> (X</span>2<span> -X</span>1<span>) (Y2 -Y1)=0
SO, the distance is just the difference in the x values.
distance = 5 --6 = 11

</span>
7 0
3 years ago
Given: circle k(O), m AM=125°, m EF=31°, m∠MAF=75°. Find: m∠AME
Bess [88]

Answer:

58°

Step-by-step explanation:

If m AM=125°, then m∠AOM=125°.

If m∠MAF=75°, then m∠MOF=150° (because central angle MOF subtends on the same arc as inscribed angle MAF).

Thus,

m∠FOA=360°-150°-125°=85°.

If mEF=31°, then m∠EOF=31° (as central angle subtended on the arc EF).

Hence,

m∠EOA=m∠EOF+m∠FOA=31°+85°=116°.

Angle EOA is central angle subtended on arc EA, angle AME is inscribed angle subtended on arc AE, thus

m∠AME=1/2m∠EOA=58°.

7 0
3 years ago
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