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Arisa [49]
3 years ago
13

Check answer now asap

Mathematics
1 answer:
kakasveta [241]3 years ago
7 0

Answer:

∠ A ≅ ∠ E

∠ K = 60°

Step-by-step explanation:

Part 1.

Given that ABCD ≅ EFGH.

Whenever a quadrilateral is congruent to another, and we write the congruency in symbol, then the order of congruency with respect to angles and sides are maintained in the symbol.

That means if ABCD ≅ EFGH, then ∠ A ≅ ∠ E, ∠ B ≅ ∠ F and so on.

Therefore, in this case, ∠ A ≅ ∠ E (Answer) {It is also shown in the diagrams}

Part 2.

Given that, Δ EFG ≅ Δ KLM

Hence, ∠ E ≅ ∠K, ∠ F ≅ ∠ L, and ∠ G ≅ ∠ M

It is also given that ∠ F = 35° and ∠ G = 85°

So, ∠ E = 180° - ∠ F - ∠ G = 180° - 85° - 35° = 60°

Since, ∠ E = ∠ K,

So, ∠ K = 60°. (Answer)

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Type the integer that makes the following subtraction sentence true:<br> -57 - ? = -151
posledela

Answer:

94

Step-by-step explanation:

Its pretty simple,

The easiest way to find the missing number is to do 151-57, which = 94

8 0
3 years ago
Doris made 80 cookies this week. She baked 4 less than three times the total she made last week. Which equation, when solved for
sertanlavr [38]

Answer:

The equation is  3x-4=80

The value of x is 28 cookies

Step-by-step explanation:

Let

x ---->  the number of cookies she baked last week

we know that

The number of cookies she baked last week multiplied by 3 minus 4 must be equal to 80 cookies

so

The linear equation that represent this situation is

3x-4=80

solve for x

3x=80+4\\3x=84\\x=28\ cookies

3 0
3 years ago
The data set of the diameters of the metal cylinders manufactured an automatic machine has sample size of n=29, mean of x= 49.98
Lena [83]

Answer:

we say for  μ = 50.00 mm we be 95% confident that machine calibrated properly with ( 49.926757  ,  50.033243 )

Step-by-step explanation:

Given data

n=29

mean of x = 49.98 mm

S = 0.14 mm

μ = 50.00 mm

Cl = 95%

to find out

Can we be 95% confident that machine calibrated properly

solution

we know from t table

t at 95% and n -1 = 29-1 = 28 is  2.048

so now

Now for  95% CI for mean is

(x - 2.048 × S/√n  ,  x + 2.048 × S/√n )

(49.98 - 2.048 × 0.14/√29  ,  49.98 + 2.048 × 0.14/√29 )

( 49.926757  ,  50.033243 )

hence we say for  μ = 50.00 mm we be 95% confident that machine calibrated properly with ( 49.926757  ,  50.033243 )

8 0
3 years ago
Quiz scores of statistics students were obtained by a teacher and listed below. Construct a frequency distribution beginning wit
DanielleElmas [232]

Answer:

The frequency distribution is

The frequency distribution is

S/N Class limit Frequency

1 0-2 4

2 3 -4 1

3 5-6 2

4 7-8 2

5 9 - 10 0

6 11 - 12 0

The value for value upper class boundary of the fifth class is CB_5 =  10.5

Step-by-step explanation:

From the question we are told that

The lower class limit is 0

The class width is 2

The scores is 8,6,2,4,2,0,2,6,7

The frequency distribution is

S/N Class limit Frequency

1 0-2 4

2 3 -4 1

3 5-6 2

4 7-8 2

5 9 - 10 0

6 11 - 12 0

Generally the upper class boundary of the fifth class is mathematically represented as

CB_5 =  \frac{10 + 11}{2}

            CB_5 =  10.5

8 0
3 years ago
Simplify and evaluate<br><br> (6)(2^3)^2 / (2^2)^2
sveticcg [70]

Simplify and evaluate

(6)(2^3)^2 = 384

(2^2)^2 = 16

5 0
2 years ago
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