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3 years ago

7 = 1/2(x + 2) Need help finding the answer!

Mathematics

1 answer:

Stels [109]3 years ago

6 0

Answer:

x=12

Step-by-step explanation:

1/2*x =1/2x

1/2*2 =1

you need to get the constants on one side so subtract 1 on all sides to get 6=1/2x, then divide by 1/2 on all sides to get x= 12

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Given g(x) = 2(x – 5), determine x for g(x) = -30.

GREYUIT [131]

When g(x) = -30, x = -10
To solve, make -30 equal to 2(x-5):
-30 = 2(x-5)
-30 = 2x -10
-20 = 2x
-10 = x

5 0

3 years ago

During one waiters shift he delivered 13 appetizers, 17 entrées, and 10 desserts what percentage of the dishes he delivered were

Triss [41]

Question

During one waiters shift he delivered 13 appetizers, 17 entrées, and 10 desserts what percentage of the dishes he delivered were: A. Desserts B. Appetizers C. entrees

Answer:

A)Desserts = 25 %

B)Appetizers = 32.5 %

C)Entrées = 32.5 %

Step-by-step explanation:

Given:

Number of appetisers = 13

Number of entrees = 17

Number of desserts = 10

A. percentage of the dishes he delivered were Desserts

Percentage of Desserts = $\frac{\text{ Number of dessert}}{\text{ total number of dishes}} \times 100$

Total number of dishes = 13 + 17 + 10 = 40

Percentage of Desserts = $\frac{10}{40} \times 100$

Percentage of Desserts = $0.25 \times 100$

Percentage of Desserts = 25 %

B. percentage of the dishes he delivered were Appetizers 

Percentage of Desserts = $\frac{\text{ Number of Appetizers }}{\text{ total number of dishes}} \times 100$

Percentage of Desserts = $\frac{13}{40} \times 100$

Percentage of Desserts = $0.325 \times 100$

Percentage of Desserts = 32.5 %

C. percentage of the dishes he delivered were Entrées 

Percentage of Desserts = $\frac{\text{ Number of Entrees }}{\text{ total number of dishes}} \times 100$

Percentage of Desserts = $\frac{17}{40} \times 100$

Percentage of Desserts = $0.425 \times 100$

Percentage of Desserts = 42.5 %

7 0

3 years ago

What is the measure of L ABD?

fenix001 [56]

Answer:

$\large\boxed{39^o}$

Step-by-step explanation:

$m\angle ABC=90^o\\\\m\angle ABC=m\angle ABD+m\angle DBC\\\\m\angle DBC=51^o\\\\\text{Substitute:}\\\\m\angle ABD+51^o=90^o\qquad\text{subtract }\ 51^o\ \text{from both sides}\\\\m\angle ABD=39^o$