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dusya [7]
3 years ago
7

$34,589 discounted 8%

Mathematics
2 answers:
Otrada [13]3 years ago
7 0

Answer: 2767.12 !!!!!!!!!!!!!!!!

Tcecarenko [31]3 years ago
6 0
The Answer is 2767.12
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Last year the manager of a health care facility served 36,500 dinners. The manager estimates an increase of 5% in the number of
EastWind [94]

Answer:

38,325 dinners.

Step-by-step explanation:

We have been given that last year the manager of a health care facility served 36,500 dinners. The manager estimates an increase of 5% in the number of dinners to be served next year.

The estimated number of dinners next year would be 36,500 plus 5% of 36,500.

36,500+(\frac{5}{100}\times 36,500)

36,500+(0.05\times 36,500)

36,500+1,825

38,325

Therefore, 38,325 dinners will be served next year.

6 0
4 years ago
Find the slope of the line 7x-4y=10
Harman [31]

Answer:

The slope is m=(7/4)

Step-by-step explanation:

we know that

The equation of the line into slope intercept form is equal to

y=mx+b

where

m is the slope

b is the y-coordinate of the y-intercept

In this problem we have

7x-4y=10

so

Convert to slope intercept form

Isolate the variable y

4y=7x-10

y=(7/4)x-(10/4)

y=(7/4)x-(5/2)

The slope is m=(7/4)

5 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
4 years ago
Awner this asap pls my cousins homework more coming up
mixas84 [53]
The answer is option 1
3 0
3 years ago
Square NOPQ is dilated by a scale factor of to form square N'O'P'Q'. Side N'O' measures 4. What is the measure of side NO?
KIM [24]
\begin{gathered} \text{NOPQ was dilated to N'O'P'Q'} \\ \text{The scale factor is }\frac{2}{3} \\ N^{\prime}O^{\prime}=4 \\ N^{\prime}O^{\prime}=\frac{2}{3}NO \\ 4=\frac{2}{3}NO \\ \text{cross multiply} \\ 12=2NO \\ NO=\frac{12}{2} \\ NO=6 \end{gathered}

4 0
1 year ago
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