Please immediate help i just need the answer

Baker made 5 pounds of icing. He used 4/9 of the icing to decorate. How much did he have left over?

Elena L [17]

If the baker made 5 pounds of icing and used 4/9 of the icing to make decorations, then he used:

5 * 4/9 =
= 20/9 =
= 2 2/9 [of icing]

To find out how much he has left over, we just need to substract 2 2/9 from 5:

5 - 2 2/9 =
= 3 - 2/9 =
= 2 7/9

Answer: The baker has 2 7/9 of the icing left.

6 0

2 years ago

The health department estimates that eight vials of malaria Serum will treat 100 people. at this rate, how many vials are requir

lara [203]

Using butterfly method, it is 14 vials

5 0

2 years ago

Jenny and Natalie are selling cheesecakes for a school fundraiser. Customers can buy chocolate cakes and vanilla cakes. Jenny so

IrinaVladis [17]

The cost of 1 chocolate cake is $ 6 and cost of 1 vanilla cake is $ 7

Solution:

Let "c" be the cost of 1 chocolate cake

Let "v" be the cost of 1 vanilla cake

Jenny sold 14 chocolate cakes and 5 vanilla cakes for 119 dollars

Therefore, we can frame a equation as:

14 x cost of 1 chocolate cake + 5 x cost of 1 vanilla cake = 119

$14 \times c + 5 \times v=119$

14c + 5v = 119 ------- eqn 1

Natalie sold 10 chocolate cakes and 10 vanilla cakes for 130 dollars

Therefore, we can frame a equation as:

10 x cost of 1 chocolate cake + 10 x cost of 1 vanilla cake = 130

$10 \times c + 10 \times v = 130$

10c + 10v = 130 -------- eqn 2

Let us solve eqn 1 and eqn 2

Multiply eqn 1 by 2

28c + 10v = 238 ------ eqn 3

Subtract eqn 2 from eqn 3

28c + 10v = 238

10c + 10v = 130

( - ) --------------------------

18c = 108

c = 6

Substitute c = 6 in eqn 1

14(6) + 5v = 119

84 + 5v = 119

5v = 119 - 84

5v = 35

v = 7

Thus cost of 1 chocolate cake is $ 6 and cost of 1 vanilla cake is $ 7

8 0

2 years ago

Please help answer these questions. My teacher said they were really easy but I just don't understand. Will mark brainliest !!!

kodGreya [7K]

Answer:

1. A = 59

2. A = 43

Step-by-step explanation:

If we have a right triangle we can use sin, cos and tan.

sin = opp/ hypotenuse

cos= adjacent/ hypotenuse

tan = opposite/ adjacent

For the first problem, we know the opposite and adjacent sides to angle A

tan A = opposite/ adjacent

tan A = 8.8 / 5.2

Take the inverse of each side

tan ^-1 tan A = tan ^-1 (8.8/5.2)

A = 59.42077313

To the nearest degree

A = 59 degrees

For the second problem, we know the adjacent side and the hypotenuse to angle A

cos A = adjacent/hypotenuse

cos A = 15.3/21

Take the inverse of each side

cos ^-1 cos A = cos ^-1 (15.3/21)

A = 43.23323481

To the nearest degree

A = 43 degrees

6 0

3 years ago

To help solve the trigonometric inequality sec(x/3)+4>2-sec(x/3) which two equations could you graph

geniusboy [140]

Answer:

b. y=sec(x/3) and y=-1

Step-by-step explanation:

sec(x/3) + 4 > 2 − sec(x/3)

2 sec(x/3) > -2

sec(x/3) > -1

Graph y = sec(x/3) and y = -1.

6 0

3 years ago