All categories

2 years ago

What are m and b in the linear equation y= 5 + 1x

Mathematics

2 answers:

hichkok12 [17]2 years ago

8 0

Answer:

m=1 and b=5

Step-by-step explanation:

y=mx+b = y= b+mx

joja [24]2 years ago

6 0

Answer:

m=1

y= 5

Step-by-step explanation:

since y=mx+b

y=1x+5

You might be interested in

Can someone solve this? and tell me how?

maksim [4K]

Answer:

1/3

Step-by-step explanation:

When working with balanced expressions (stuff on both sides of the equal sign), "what you do to one side, you do to the other", which keeps it balanced.

The first thing we notice is the exponent 1/4, which is one both sides, so we can get rid of it on both sides by using the reverse operation.

The reverse of exponents is square root.

$(4x + 10)^{\frac{1}{4}} = (9 + 7x)^{\frac{1}{4}}\\\sqrt[\frac{1}{4}]{(4x + 10)^{\frac{1}{4}}} = \sqrt[\frac{1}{4}]{(9 + 7x)^{\frac{1}{4}}}\\\\4x + 10 = 9 + 7x$

Isolate x to solve. Separate the variables and non-variables.

4x + 10 = 9 + 7x

4x - 4x + 10 = 9 + 7x - 4x Subtract 4x from both sides

10 = 9 + 3x

10 - 9 = 9 - 9 + 3x Subtract 9 from both sides

1 = 3x Divide both sides by 3 to isolate x

x = 1/3 Answer

8 0

2 years ago

Hellooooo everyone its donald trump!! ihave a question?What led to the decline of the sugar business in the West Indies?

lbvjy [14]

D) doctors told peopel thag suger would damage their health

5 0

3 years ago

Read 2 more answers

Angelo is making a rectangular floor for a clubhouse with an area of 84 square feet. The length if each side of the floor is a w

Alexxandr [17]

Answer:

L=1 ft and B=84 ft

L=2 ft and B=42 ft

L=4ft and B=21ft

L=6ft and B=14 ft

L=7ft and B=12ft

L=12 ft and B=7ft

L=14ft and B=6 ft

L=21 ft and B=4 ft

L=42 ft and B=2 ft

L=84 ft and B=1 ft

Step-by-step explanation:

We are given that

Area of rectangular floor=84 square feet

We have to find the possible length and width of for Angelo's clubhouse.

Area of rectangle= $length\times breadth$

Using the formula

Area of rectangular floor for clubhouse=84 square feet

$Length(L)\times breadth(B)=84$

Factor of 84 are

1,2,4,6,7,12,14,21,42,84

Therefore, possible dimension of rectangular floor

$1ft\times 84ft,2ft\times 42ft,4ft\times 21ft,6ft\times 14ft,7ft\times 12ft,12ft\times 7ft,14ft\times 6ft,21ft\times 4ft,42ft\times 2ft,84ft\times 1ft$

L=1 ft and B=84 ft

L=2 ft and B=42 ft

L=4ft and B=21ft

L=6ft and B=14 ft

L=7ft and B=12ft

L=12 ft and B=7ft

L=14ft and B=6 ft

L=21 ft and B=4 ft

L=42 ft and B=2 ft

L=84 ft and B=1 ft

5 0

2 years ago

I think number 9 is b or c, i don't know what 10 is. please help me!!!

scoray [572]

10 is letter b. I hope this helps!!!

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7B%281-x%5E%7B2%7D%20%29%5E%7B3%2F2%7D%20%7D%20%5C%2C%20dx" id="TexFo

Ludmilka [50]

Answer: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(a) + 2a(1 - a^2)^\Big{\frac{3}{2}} + 3a\sqrt{1 - a^2}}{8} - \frac{3arcsin(b) + 2b(1 - b^2)^\Big{\frac{3}{2}} + 3b\sqrt{1 - b^2}}{8}$ General Formulas and Concepts:

Pre-Calculus

Trigonometric Identities

Calculus

Differentiation

Derivatives
Derivative Notation

Integration

Integrals
Definite/Indefinite Integrals
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]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

U-Substitution

Trigonometric Substitution

Reduction Formula: $\displaystyle \int {cos^n(x)} \, dx = \frac{n - 1}{n}\int {cos^{n - 2}(x)} \, dx + \frac{cos^{n - 1}(x)sin(x)}{n}$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution (trigonometric substitution).

Set u: $\displaystyle x = sin(u)$
[u] Differentiate [Trigonometric Differentiation]: $\displaystyle dx = cos(u) \ du$
Rewrite u: $\displaystyle u = arcsin(x)$

Step 3: Integrate Pt. 2

[Integral] Trigonometric Substitution: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos(u)[1 - sin^2(u)]^\Big{\frac{3}{2}} \, du$
[Integrand] Rewrite: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos(u)[cos^2(u)]^\Big{\frac{3}{2}} \, du$
[Integrand] Simplify: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos^4(u)} \, du$
[Integral] Reduction Formula: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{4 - 1}{4}\int \limits^a_b {cos^{4 - 2}(x)} \, dx + \frac{cos^{4 - 1}(u)sin(u)}{4} \bigg| \limits^a_b$
[Integral] Simplify: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4}\int\limits^a_b {cos^2(u)} \, du$
[Integral] Reduction Formula: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg|\limits^a_b + \frac{3}{4} \bigg[ \frac{2 - 1}{2}\int\limits^a_b {cos^{2 - 2}(u)} \, du + \frac{cos^{2 - 1}(u)sin(u)}{2} \bigg| \limits^a_b \bigg]$
[Integral] Simplify: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4} \bigg[ \frac{1}{2}\int\limits^a_b {} \, du + \frac{cos(u)sin(u)}{2} \bigg| \limits^a_b \bigg]$
[Integral] Reverse Power Rule: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4} \bigg[ \frac{1}{2}(u) \bigg| \limits^a_b + \frac{cos(u)sin(u)}{2} \bigg| \limits^a_b \bigg]$
Simplify: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3cos(u)sin(u)}{8} \bigg| \limits^a_b + \frac{3}{8}(u) \bigg| \limits^a_b$
Back-Substitute: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(arcsin(x))sin(arcsin(x))}{4} \bigg| \limits^a_b + \frac{3cos(arcsin(x))sin(arcsin(x))}{8} \bigg| \limits^a_b + \frac{3}{8}(arcsin(x)) \bigg| \limits^a_b$
Simplify: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(x)}{8} \bigg| \limits^a_b + \frac{x(1 - x^2)^\Big{\frac{3}{2}}}{4} \bigg| \limits^a_b + \frac{3x\sqrt{1 - x^2}}{8} \bigg| \limits^a_b$
Rewrite: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(x) + 2x(1 - x^2)^\Big{\frac{3}{2}} + 3x\sqrt{1 - x^2}}{8} \bigg| \limits^a_b$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(a) + 2a(1 - a^2)^\Big{\frac{3}{2}} + 3a\sqrt{1 - a^2}}{8} - \frac{3arcsin(b) + 2b(1 - b^2)^\Big{\frac{3}{2}} + 3b\sqrt{1 - b^2}}{8}$