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sladkih [1.3K]
2 years ago
5

Answer quick please!

Mathematics
2 answers:
8_murik_8 [283]2 years ago
8 0

Answer:

y = x/2 + 5

Step-by-step explanation:

y = x/2 + 5

⇒ y = (1/2)x + 5

which satisfies the linear equation y = mx + b

where m is the slope and b is the y-intercept

eimsori [14]2 years ago
7 0
Answer is B hope I helped!
You might be interested in
PLEASE HELP I NEED THIS BADLY!!! ANSWER WILL GET BRAINLIEST!
earnstyle [38]

Answer:

A': (4,-4)

B': (3,-8)

C': (3,-5)

Step-by-step explanation:

Translation

A (5,2)---> (4,4)

B(4,6)---> (3,8)

C(4,3)---> (3,5)

Reflection over x-axis= (x,y) to (x,-y)

A': (4,-4)

B': (3,-8)

C': (3,-5)

Hope this helps!

Brainliest?

Have a good day!

7 0
4 years ago
Read 2 more answers
(つ ◕.◕ )つ (Please Help Asap, Tysm!)
ankoles [38]

Answer:

░░ ░░░░▄░░░░░░░░░ ▄ ░░░░░░░░▌▒█░░░░░░░░░░░▄▀▒▌ ░░░░░░░░▌▒▒█░░░░░░░░▄▀▒▒▒▐ ░░░░░░░▐▄▀▒▒▀▀▀▀▄▄▄▀▒▒▒▒▒▐ ░░░░░▄▄▀▒░▒▒▒▒▒▒▒▒▒█▒▒▄█▒▐ ░░░▄▀▒▒▒░░░▒▒▒░░░▒▒▒▀██▀▒▌ ░░▐▒▒▒▄▄▒▒▒▒░░░▒▒▒▒▒▒▒▀▄▒▒▌ ░░▌░░▌█▀▒▒▒▒▒▄▀█▄▒▒▒▒▒▒▒█▒▐ ░▐░░░▒▒▒▒▒▒▒▒▌██▀▒▒░░░▒▒▒▀▄▌ ░▌░▒▄██▄▒▒▒▒▒▒▒▒▒░░░░░░▒▒▒▒▌ ▌▒▀▐▄█▄█▌▄░▀▒▒░░░░░░░░░░▒▒▒▐ ▐▒▒▐▀▐▀▒░▄▄▒▄▒▒▒▒▒▒░▒░▒░▒▒▒▒▌ ▐▒▒▒▀▀▄▄▒▒▒▄▒▒▒▒▒▒▒▒░▒░▒░▒▒▐ ░▌▒▒▒▒▒▒▀▀▀▒▒▒▒▒▒░▒░▒░▒░▒▒▒▌ ░▐▒▒▒▒▒▒▒▒▒▒▒▒▒▒░▒░▒░▒▒▄▒▒▐ ░░▀▄▒▒▒▒▒▒▒▒▒▒▒░▒░▒░▒▄▒▒▒▒▌ ░░░░▀▄▒▒▒▒▒▒▒▒▒▒▄▄▄▀▒▒▒▒▄▀ ░░░░░░▀▄▄▄▄▄▄▀▀▀▒▒▒▒▒▄▄—

Step-by-step explanation:

░░ ░░░░▄░░░░░░░░░ ▄ ░░░░░░░░▌▒█░░░░░░░░░░░▄▀▒▌ ░░░░░░░░▌▒▒█░░░░░░░░▄▀▒▒▒▐ ░░░░░░░▐▄▀▒▒▀▀▀▀▄▄▄▀▒▒▒▒▒▐ ░░░░░▄▄▀▒░▒▒▒▒▒▒▒▒▒█▒▒▄█▒▐ ░░░▄▀▒▒▒░░░▒▒▒░░░▒▒▒▀██▀▒▌ ░░▐▒▒▒▄▄▒▒▒▒░░░▒▒▒▒▒▒▒▀▄▒▒▌ ░░▌░░▌█▀▒▒▒▒▒▄▀█▄▒▒▒▒▒▒▒█▒▐ ░▐░░░▒▒▒▒▒▒▒▒▌██▀▒▒░░░▒▒▒▀▄▌ ░▌░▒▄██▄▒▒▒▒▒▒▒▒▒░░░░░░▒▒▒▒▌ ▌▒▀▐▄█▄█▌▄░▀▒▒░░░░░░░░░░▒▒▒▐ ▐▒▒▐▀▐▀▒░▄▄▒▄▒▒▒▒▒▒░▒░▒░▒▒▒▒▌ ▐▒▒▒▀▀▄▄▒▒▒▄▒▒▒▒▒▒▒▒░▒░▒░▒▒▐ ░▌▒▒▒▒▒▒▀▀▀▒▒▒▒▒▒░▒░▒░▒░▒▒▒▌ ░▐▒▒▒▒▒▒▒▒▒▒▒▒▒▒░▒░▒░▒▒▄▒▒▐ ░░▀▄▒▒▒▒▒▒▒▒▒▒▒░▒░▒░▒▄▒▒▒▒▌ ░░░░▀▄▒▒▒▒▒▒▒▒▒▒▄▄▄▀▒▒▒▒▄▀ ░░░░░░▀▄▄▄▄▄▄▀▀▀▒▒▒▒▒▄▄—

4 0
3 years ago
Read 2 more answers
Hi, how do we do this question?​
Nutka1998 [239]

Answer:

\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{-2(ln|3x + 1| - 3x)}{9} + C

General Formulas and Concepts:

<u>Algebra I</u>

  • Terms/Coefficients
  • Factoring

<u>Algebra II</u>

  • Polynomial Long Division

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

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

Derivative Property [Addition/Subtraction]:                                                         \displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]  

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Integration

  • Integrals
  • Integration Constant C
  • Indefinite Integrals

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

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

Logarithmic Integration

U-Substitution

Step-by-step explanation:

*Note:

You could use u-solve instead of rewriting the integrand to integrate this integral.

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle \int {\frac{2x}{3x + 1}} \, dx

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

  1. [Integrand] Rewrite [Polynomial Long Division (See Attachment)]:           \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \int {\bigg( \frac{2}{3} - \frac{2}{3(3x + 1)} \bigg)} \, dx
  2. [Integral] Rewrite [Integration Property - Addition/Subtraction]:               \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \int {\frac{2}{3}} \, dx - \int {\frac{2}{3(3x + 1)}} \, dx
  3. [Integrals] Rewrite [Integration Property - Multiplied Constant]:               \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}\int {} \, dx - \frac{2}{3}\int {\frac{1}{3x + 1}} \, dx
  4. [1st Integral] Reverse Power Rule:                                                               \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{3}\int {\frac{1}{3x + 1}} \, dx

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

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

  1. Set <em>u</em>:                                                                                                             \displaystyle u = 3x + 1
  2. [<em>u</em>] Differentiate [Basic Power Rule]:                                                             \displaystyle du = 3 \ dx

<u>Step 4: Integrate Pt. 3</u>

  1. [Integral] Rewrite [Integration Property - Multiplied Constant]:                 \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}\int {\frac{3}{3x + 1}} \, dx
  2. [Integral] U-Substitution:                                                                               \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}\int {\frac{1}{u}} \, du
  3. [Integral] Logarithmic Integration:                                                               \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}ln|u| + C
  4. Back-Substitute:                                                                                            \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}ln|3x + 1| + C
  5. Factor:                                                                                                           \displaystyle \int {\frac{2x}{3x + 1}} \, dx = -2 \bigg( \frac{1}{9}ln|3x + 1| - \frac{x}{3}  \bigg) + C
  6. Rewrite:                                                                                                         \displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{-2(ln|3x + 1| - 3x)}{9} + C

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

Unit: Integration

Book: College Calculus 10e

8 0
3 years ago
Flaming BBQ Restaurant makes a dipping sauce with 9\text{ mL}9 mL9, space, m, L of hot sauce for every 666 ounces of barbecue sa
ahrayia [7]

Answer:

3 mL of hot sauce mixed with 2 ounces of barbecue sauce

45 mL of hot sauce mixed with 30 ounces of barbecue sauce

12 mL of hot sauce mixed with 8 ounces of barbecue sauce

Step-by-step explanation:

we know that

A restaurant makes a dipping sauce with

\frac{9}{6}\ \frac{mL\ hot\ sauce}{oz\ barbecue\ sauce}

Simplify

1.5\ \frac{mL\ hot\ sauce}{oz\ barbecue\ sauce}

<u><em>Verify each case</em></u>

case A) we have

6 mL of hot sauce mixed with 10 ounces of barbecue sauce

so

\frac{6}{10}\ \frac{mL\ hot\ sauce}{oz\ barbecue\ sauce}

simplify

0.6\ \frac{mL\ hot\ sauce}{oz\ barbecue\ sauce}

Compare

0.6 \neq 1.5

therefore

This mixture is not the same that as Flaming BBQ's dipping sauce

case B) we have

3 mL of hot sauce mixed with 2 ounces of barbecue sauce

so

\frac{3}{2}\ \frac{mL\ hot\ sauce}{oz\ barbecue\ sauce}

simplify

1.5\ \frac{mL\ hot\ sauce}{oz\ barbecue\ sauce}

Compare

1.5 =1.5

therefore

This mixture is the same that as Flaming BBQ's dipping sauce

case C) we have

45 mL of hot sauce mixed with 30 ounces of barbecue sauce

so

\frac{45}{30}\ \frac{mL\ hot\ sauce}{oz\ barbecue\ sauce}

simplify

1.5\ \frac{mL\ hot\ sauce}{oz\ barbecue\ sauce}

Compare

1.5=1.5

therefore

This mixture is the same that as Flaming BBQ's dipping sauce

case D) we have

24 mL of hot sauce mixed with 18 ounces of barbecue sauce

so

\frac{24}{18}\ \frac{mL\ hot\ sauce}{oz\ barbecue\ sauce}

simplify

1.33\ \frac{mL\ hot\ sauce}{oz\ barbecue\ sauce}

Compare

1.33 \neq 1.5

therefore

This mixture is not the same that as Flaming BBQ's dipping sauce

case E) we have

12 mL of hot sauce mixed with 8 ounces of barbecue sauce

so

\frac{12}{8}\ \frac{mL\ hot\ sauce}{oz\ barbecue\ sauce}

simplify

1.5\ \frac{mL\ hot\ sauce}{oz\ barbecue\ sauce}

Compare

1.5=1.5

therefore

This mixture is the same that as Flaming BBQ's dipping sauce

6 0
3 years ago
A dairy farm produced 254.8 liters
Ivan
4/254.8 to find the answer

Answer=63.7
6 0
3 years ago
Read 2 more answers
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