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Yuliya22 [10]
3 years ago
14

Choose the answer.

Mathematics
2 answers:
Tasya [4]3 years ago
6 0
1. B she started with 36 photographs,
22 = b - 18 + 13 - 9
b = 22 + 18 - 13 + 9
b = 36

I translated it into an equation, so I would say C, but since B is the only one with the right answer...

2. A
78 - 5 = 73
73 × 365 = 26645

3. D

4. D

5. A
40 × 86 = 3440
Andrej [43]3 years ago
6 0
1. B
2. A
3. D
4. D
5. A
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lozanna [386]

Answer:

A.

Step-by-step explanation:

as u can see,the other acute angle is 43° and the other is also an acute angle and acute angles must not surpass until 80° meaning, the other acute angle is 47°

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Oksanka [162]

Answer:

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Step-by-step explanation:

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3 years ago
Enter your answer and show all the steps that you use to solve this problem in the space provided. Find the values of x and y.
satela [25.4K]

Answer : The value of x and y is, 90^o and 43^oC

Step-by-step explanation :

First we have to calculate the angle B.

As we know that,

The given triangle is an isosceles triangle in which the angles opposite to equal sides are always equal.

Thus, \angle B=\angle C

Given : \angle C=47^o

\angle B=\angle C=47^o

\angle B=47^o

Now have to calculate the angle A.

As we know that the sum of interior angles of a triangle is equal to 180^o.

\angle A+\angle B+\angle C=180^o

\angle A+47^o+47^o=180^o

\angle A+94^o=180^o

\angle A=180^o-94^o

\angle A=86^o

Now we have to determine the angle y.

As we know that, line AD is an angle bisector. That means, it divides into two equal angles. So,

\angle y=\frac{\angle A}{2}

\angle y=\frac{86^o}{2}

\angle y=43^o

Now we have to determine the angle x.

As we know that the sum of interior angles of a triangle is equal to 180^o.

\angle y+\angle B+\angle x=180^o

43^oc+47^o+\angle x=180^o

\angle x+90^o=180^o

\angle x=180^o-90^o

\angle x=90^o

Thus, the value of x and y is, 90^o and 43^oC

6 0
3 years ago
100 POINTS<br><br> PLEASE PROVIDE STEPS<br> THANK YOU!!
kolbaska11 [484]

Answer:

Local minimum at x = 0.

Step-by-step explanation:

Local minimums occur when g'(x) = 0 and g"(x) > 0.

Local maximums occur when g'(x) = 0 and g"(x) < 0.

Set g'(x) equal to 0 and solve:

0 = 2x (x − 1)² (x + 1)²

x = 0, 1, or -1

Evaluate g"(x) at each point:

g"(0) = 2

g"(1) = 0

g"(-1) = 0

There is a local minimum at x = 0.

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