8.1g of sugar is needed for every cake made. How much sugar is needed for 6 cakes?

Mathematics

1 answer:

Lerok [7]3 years ago

8 0

Answer:

48.6

Step-by-step explanation:

If you use 8.1g of sugar for 1 cake then 6 cakes will be 48.6g of sugar

Just do 8.1*6 and you will get 48.6

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Explain how to use proportional reasoning to find 35% of 600

Sauron [17]

$\frac{35}{100} = \frac{x}{600}$

This is the equation we would use for this problem.

Now when we cross multiply, we get:
35(600)=100x

Simplify:
21000=100x

Divide each side by 100 to get x b itself:
210=x

So, as you can see, 210 is 35% of 600

8 0

3 years ago

Sony brother is 8 year older than four fifth her age. if her brothers age is 32 years what is sony

madam [21]

Answer:

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3 0

3 years ago

Use the Fundamental Theorem of Calculus to find the "area under curve" of

lozanna [386]

Answer:

$\displaystyle A = 300$

General Formulas and Concepts:

Calculus

Integrals

Definite Integrals
Area under the curve
Integration Constant C

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Area of a Region Formula: $\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx$

Step-by-step explanation:

Step 1: Define

Identify

f(x) = 6x + 19

Interval [12, 15]

Step 2: Find Area

Substitute in variables [Area of a Region Formula]: $\displaystyle A = \int\limits^{15}_{12} {(6x + 19)} \, dx$
[Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle A = \int\limits^{15}_{12} {6x} \, dx + \int\limits^{15}_{12} {19} \, dx$
[Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle A = 6\int\limits^{15}_{12} {x} \, dx + 19\int\limits^{15}_{12} {} \, dx$
[Integrals] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle A = 6(\frac{x^2}{2}) \bigg| \limits^{15}_{12} + 19(x) \bigg| \limits^{15}_{12}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle A = 6(\frac{81}{2}) + 19(3)$
Simplify: $\displaystyle A = 300$