Ask question

Ask question

All categories

3 years ago

10

Rashida uses 8 cups of tomatoes and 3 cups of onions to make salsa. How many cups of onions should rashida use if she uses only

4 cups of tomatoes

2 answers:

prisoha [69]3 years ago

5 0

Since half of 8 is for u should cut 3 in half which would give u 1.5

geniusboy [140]3 years ago

3 0

1 1/2 cups
hope this helps:)

You might be interested in

What is dilation and can someone explain in a easy way

gogolik [260]

dilation is when a shape changes size, from small to big or big to small.

4 0

3 years ago

Read 2 more answers

If anyone knows about definite integrals for calculus then please I request help! I

kicyunya [14]

Answer:

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals

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

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

U-Substitution

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx$

<u>Step 2: Integrate Pt. 1</u>

<em>Identify variables for u-substitution.</em>

Set <em>u</em>: $\displaystyle u = 4x^{-2}$
[<em>u</em>] Differentiate [Basic Power Rule, Derivative Properties]: $\displaystyle du = \frac{-8}{x^3} \ dx$
[Bounds] Switch: $\displaystyle \left \{ {{x = 9 ,\ u = 4(9)^{-2} = \frac{4}{81}} \atop {x = 5 ,\ u = 4(5)^{-2} = \frac{4}{25}}} \right.$

<u>Step 3: Integrate Pt. 2</u>

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^9_5 {\frac{-8}{x^3}e^\big{4x^{-2}}} \, dx$
[Integral] U-Substitution: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{4}{81}}_{\frac{4}{25}} {e^\big{u}} \, du$
[Integral] Exponential Integration: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}(e^\big{u}) \bigg| \limits^{\frac{4}{81}}_{\frac{4}{25}}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8} \bigg( e^\Big{\frac{4}{81}} - e^\Big{\frac{4}{25}} \bigg)$
Simplify: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

4 0

3 years ago

Given the function f(x) = 2/3 * x - 5 , evaluate f(9)

ch4aika [34]

Answer:

1

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Need help with working perimeter and area please

SVETLANKA909090 [29]

Answer:

Area = 228 m²

Perimeter = 60 m

Step-by-step explanation:

The figure given shows a rectangle that has a cut triangular portion.

✔️Area of the figure = area of rectangle - area of the triangular cut portion

= L*W + ½*bh

Where,

L = 20 m

W = 12 m

b = 20 - (8 + 8) = 4 m

h = 6 m

Plug in the values

Area = 20*12 - ½*4*6

Area = 240 - 12

Area = 228 m²

✔️Perimeter = perimeter of rectangle - base of the triangular cut portion

= 2(L + W) - b

L = 20 m

W = 12 m

b = b = 20 - (8 + 8) = 4 m

Plug in the values

Perimeter = 2(20 + 12) - 4

= 2(32) - 4

= 64 - 4

Perimeter = 60 m

5 0

3 years ago

Draw to show regrouping. Write how many tens and ones in the sum. Write the sum. This has me so confused.

olya-2409 [2.1K]

I am not sure, but we're you going to put up a sum.

7 0

3 years ago