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3 years ago

Help! Will give brainliest and 10 points!

Mathematics

1 answer:

krok68 [10]3 years ago

3 0

Answer:

-3a^2 - 6a - 10

Step-by-step explanation:

You are subtracting 10a^2 + 6a + 2 from 7a^2 - 8.

You can set it up like the problem shows you, and subtract each term in the second line form a like term in the top line.

7a^2 -8

(-) 10a^2 + 6a + 2

-----------------------------

-3a^2 - 6a - 10

Answer: -3a^2 - 6a - 10

You might be interested in

4x-5y=-15 7x-2y=21 X= Y=

Helen [10]

Answer:

The values of x and y to the given equations are x=5 and y=7

Step-by-step explanation:

Given equations are $4x-5y=-15\hfill (1)$

$7x-2y=21\hfill (2)$

To solve the given equations by elimination method :

Multiply the equation (1) into 2 we get

$8x-10y=-30\hfill (3)$

Multiply the equation (2) into 5 we get

$35x-10y=105\hfill (4)$

Now subtracting the equations (3) and (4) we get

$8x-10y=-30$

$35x-10y=105$

_________________

$-27x=-135$

$x=\frac{135}{27}$

Therefore x=5

Now substitute the value of x=5 in equation (1) we get

4(5)-5y=-15

20-5y=-15

-5y=-15-20

-5y=-35

$y=\frac{35}{5}$

Therefore y=7

The values are x=5 and y=7

4 0

3 years ago

I need help. Also need the answer in step by step form

Nutka1998 [239]

Answer:

x=2

Step-by-step explanation:

f(x)=-3x+1, f(x)=-5

-5=-3x+1, 3x=6, x=2. Answered by gauthmath

7 0

3 years ago

Read 2 more answers

Janna has 34 pennies. Mandy has 48 pennies. Mandy says they have 72

krok68 [10]

Answer:

Mandy should add the tens before she adds the ones.

Step-by-step explanation:

There are no other good answers and the correct answer is 82 pennies.

3 0

3 years ago

Read 2 more answers

The system shown has _____ solution(s).

loris [4]

This system has one solution. Here are my steps to solving this problem:

y=x+1

2(x+1) - x = 2

2x +2 - x = 2

2x - x + 2 = 2

x + 2 = 2
- 2 - 2
----------------
x = 0

y = 0 + 1
y = 1

I hope this helps you!

5 0

3 years ago

Read 2 more answers

Calculus 2. Please help

Anarel [89]

Answer:

$\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}}} \, dx = \infty$

General Formulas and Concepts:

Algebra I

Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Calculus

Limits

Right-Side Limit: $\displaystyle \lim_{x \to c^+} f(x)$

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integrals

Definite Integrals

Integration Constant C

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

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

U-Substitution

U-Solve

Improper Integrals

Exponential Integral Function: $\displaystyle \int {\frac{e^x}{x}} \, dx = Ei(x) + C$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx$

Step 2: Integrate Pt. 1

[Integral] Rewrite [Exponential Rule - Rewrite]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \int\limits^1_0 {\frac{e^{-x^2}}{x} \, dx$
[Integral] Rewrite [Improper Integral]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \int\limits^1_a {\frac{e^{-x^2}}{x} \, dx$

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

Set: $\displaystyle u = -x^2$
Differentiate [Basic Power Rule]: $\displaystyle \frac{du}{dx} = -2x$
[Derivative] Rewrite: $\displaystyle du = -2x \ dx$

Rewrite u-substitution to format u-solve.

Rewrite du: $\displaystyle dx = \frac{-1}{2x} \ dx$

Step 4: Integrate Pt. 3

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} -\int\limits^1_a {-\frac{e^{-x^2}}{x} \, dx$
[Integral] Substitute in variables: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} -\int\limits^1_a {\frac{e^{u}}{-2u} \, du$
[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{1}{2}\int\limits^1_a {\frac{e^{u}}{u} \, du$
[Integral] Substitute [Exponential Integral Function]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{1}{2}[Ei(u)] \bigg| \limits^1_a$
Back-Substitute: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{1}{2}[Ei(-x^2)] \bigg| \limits^1_a$
Evaluate [Integration Rule - FTC 1]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{1}{2}[Ei(-1) - Ei(a)]$
Simplify: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{Ei(-1) - Ei(a)}{2}$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \infty$

∴ $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx$ diverges.

Topic: Multivariable Calculus

7 0

3 years ago