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3 years ago

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A gym membership charges a one-time application fee of $15, plus $32 each month. Which of the following equations could represen

t the total cost, y, for x number of months?
A.y = 32x + 15
B.y = 15x + 32
C.y = 47x
D.15y = 32x

1 answer:

avanturin [10]3 years ago

4 0

Answer:

The answer is A. y =32x + 15

Step-by-step explanation:

The reason behind this is the rate. The amount of money each month is the slope (m). The additional fee would be the b.

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if the scale drawing is 1:30,what should you do with each measurements of the drawing to get the actual dimensions. Plz help fas

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The scale will continuously go up per section, 2:60, 3:90.

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3 years ago

Simplify by combining like terms: 5x + 3x + 10x

Maurinko [17]

Answer:

18x

Step-by-step explanation:

Since all 3 are monomial terms with the same variable, we can add them together:

5x + 3x = 8x + 10x = 18x

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3 years ago

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Determine the area enclosed by y=2x+3, the x-axis and the ordinates x=3 and x=4

jok3333 [9.3K]

Answer:

$\displaystyle \int\limits^4_3 {2x + 3} \, dx = 10$

General Formulas and Concepts:
<u>Calculus</u>

Integration

Integrals

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:

<u>Step 1: Define</u>

<em>Identify.</em>

y = 2x + 3

<em>x</em>-interval [3, 4]

<em>x</em>-axis

<em>See attachment for graph.</em>

<u>Step 2: Find Area</u>

Substitute in variables [Area of a Region Formula]: $\displaystyle A = \int\limits^4_3 {2x + 3} \, dx$
[Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle A = \int\limits^4_3 {2x} \, dx + \int\limits^4_3 {3} \, dx$
[Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle A = 2 \int\limits^4_3 {x} \, dx + 3 \int\limits^4_3 {} \, dx$
[Integrals] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle A = 2 \bigg( \frac{x^2}{2} \bigg) \bigg| \limits^4_3 + 3(x) \bigg| \limits^4_3$
[Integrals] Integrate [Integration Rule - FTC 1]: $\displaystyle A = 2 \bigg( \frac{7}{2} \bigg) + 3(1)$
Simplify: $\displaystyle A = 10$

∴ the area bounded by the region y = 2x + 3, x-axis, and the coordinates x = 3 and x = 4 is equal to 10.

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Learn more about integration: brainly.com/question/26401241

Learn more about calculus: brainly.com/question/20197752

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

Unit: Integration

3 0

2 years ago

Triangles O N M and S R Q are shown. Angles O N M and S R Q are congruent. The length of side N M is 10 and the length of side S

valina [46]

Answer:

Option C.

Step-by-step explanation:

In △ONM and △SRQ,

$\angle ONM\cong \angle SRQ$

$NM=10$

$SR=20$

$NO=8$

$QR=x$

We need to find the value of x that will make △ONM similar to △SRQ by the SAS similarity theorem.

According to SAS similarity theorem, two triangle are similar if two corresponding sides in both triangles are proportional and the included angle in both are congruent.

It is given that $\angle ONM\cong \angle SRQ$ . So, both triangles are similar by SAS if

$\dfrac{NO}{NM}=\dfrac{SR}{QR}$

Substitute the given values.

$\dfrac{8}{10}=\dfrac{20}{x}$

$8\times x=20\times 10$

$8x=200$

Divide both sides by 8.

$x=\dfrac{200}{8}$

$x=25$

Therefore, the correct option is C.

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3 years ago

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If f(x) = -3x4 - 2x3 + 3x2, and g(x) = 3x4 - 4x3 + x2 then f(x) + g(x) =

Alex_Xolod [135]

I'm assuming the the numbers coming after the variables are meant to be exponents, and if I'm wrong just let me know

-3x^4 - 2x^3 + 3x^2 + 3x^4 - 4x^3 + x^2, rearrange to have like terms in order
-3x^4 +3x^4 - 2x^3 - 4x^3 +3x^2 + x^2, now simplify
0 -6x^3 + 4x^2

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3 years ago