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kolbaska11 [484]
3 years ago
7

Question 1 of 25

Mathematics
1 answer:
aleksklad [387]3 years ago
8 0
SAS i believe :) sorry if im wrong!
You might be interested in
In sunlight, a vertical yardstick casts a 1 ft shadow at the same time that a nearby tree casts a 15 ft shadow. How tall is the
nadya68 [22]

To answer this specific problem, the tree is 45 feet tall. I am hoping that this answer has satisfied your query and it will be able to help you, and if you would like, feel free to ask another question.

7 0
3 years ago
Read 2 more answers
10°<br> 55°<br> x х<br> Solve for y in the figure above.
andreyandreev [35.5K]
Y=145
Explanation:
A triangle’s angles add up to a total of 180 so you would subtract 55 from 180 and then subtract 90 because of the right angle, then you would end with x=35. Since x and y are supplementary they add to 180...so then you subtract 35 from 180 and then you get y=145.
8 0
3 years ago
10) 28n4 + 16n? – 80n² = 0<br> Factoring quadratic
Setler79 [48]

Answer:

2n(14n^2+3)=28n^4+16n I'm not sure if you meant to put the question mark between or not but I did it separately for -80^2=0 n would equal zero hope this helped

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
For which operations is the set {0, 1} closed?
velikii [3]
0 × <span>0 = 0
0 </span>× <span>1 = 0
1 </span>× <span>1 = 1

Which means multiplication is closed under {0, 1}

</span><span>1 </span>÷ <span>1 = 1
0 </span>÷ <span>1 = 0
</span>
Division is not closed under {0, 1}

1 + 1 = 2

Addition is not closed under {0, 1}

0 - 1 = -1

Subtraction is not closed under {0, 1} either



So it's only A. Multiplication which is closed under {0, 1}
5 0
3 years ago
Find the length of the tangent segment AB to the circles centered at O and O' whose radii are a and b respectively when the circ
bixtya [17]

Answer:

AB=2\sqrt{ab}

Step-by-step explanation:

From figure,

OA=a,  \quad \quad O'B=b\\\Rightarrow OO'= (a+b) \quad \quad \text{and}\quad OD=(a-b)

In triangle OO'D

  (OO')^2=(OD)^2+(O' D)^2

\Rightarrow (a+b)^2=(a-b)^2+(O' D)^2\\\Rightarrow a^2+b^2+2ab-a^2-b^2+2ab=(O' D)^2\\\Rightarrow 4ab=(O' D)^2\\\Rightarrow O'D=2\sqrt{ab} \\\Rightarrow O' D=2\sqrt{ab}=AB \quad \quad [\because O' DAB\;\; \text{is a rectangle.}]

Hence, AB=2\sqrt{ab}

6 0
3 years ago
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