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snow_lady [41]
3 years ago
9

Write an equation of the line with a slope of 2 and y-intercept of 9

Mathematics
1 answer:
spayn [35]3 years ago
7 0

Answer:

Y=2x+9 :)

Step-by-step explanation:


You might be interested in
T=21and u =4 what is \sqrt(t+U)
Aleks04 [339]

√(t + u)

√(21 + 4)

√25

5

6 0
2 years ago
Read 2 more answers
If AC equals 48 meters, what is the perimeter of the field
Alinara [238K]

Remark

A kite is constructed such that AB = BC and AD = DC. AB = sqrt( (1/2)AC + 18^2) see diagram. AD = sqrt(24^2 + 32^2)

Step One

Solve for AB

1/2 AC = 24 (AC is given as 48)

18 is a given length

AB = sqrt(24^2 + 18^2) = sqrt(576 + 324) = sqrt(900) = 30

Step Two

Find the length of AD

AD = sqrt(32^2 + 24^2) = sqrt(1024 + 576) = sqrt(1600) = 40

Step Three

Find the Perimeter.

P = 2 * 30 + 2*40 = 60 + 80 = 140

P = 140 <<<<< Answer

5 0
3 years ago
Read 2 more answers
Initially, there are 40 grams of A and 50 grams of B, and for each gram of B, 2 grams of A is used. It is observed that 15 grams
hram777 [196]

Answer:

X(16)=25.71grams

Step-by-step explanation:

let X(t) denote grams of C formed in  t mins.

For X grams of C we have:

\frac{2}{3}Xg of A and \frac{1}{3}Xg of B

Amounts of A,B remaining at any given time is expressed as:

40-\frac{2}{3}Xg of A and  50-\frac{1}{3}Xg  of B

Rate at which C is formed satisfies:

\frac{dX}{dt} \infty(40-\frac{2}{3}X)(50-\frac{1}{3}X)->\frac{dX}{dt}=k(90-X)\\\therefore \frac{dX}{(90-X)^2}=kdt->\int{\frac{dX}{(90-X)^2}} \, =\int {k} \, dt  \\\therefore \frac{1}{90-X}=kt+c->90-X=\frac{1}{kt+c}\\\\X(t)=90-\frac{1}{kt+c}

Apply the initial condition,X(0)=0 ,to the expression above

0=90-\frac{1}{c} \ \ ->c=\frac{1}{90}\\\therefore\\X(t)=90-\frac{1}{kt+\frac{1}{90}} \ \ ->X(t)=90-\frac{90}{90kt+c}

Now at X(8)=15:

15=90-\frac{90}{90\times 8k+1}  \ ->75=\frac{90}{720k+1}\\k=0.0002778

Substitute  in X(t) to get

X(t)=90-\frac{90}{0.0002778t\times 90+1}\\X(t)=90-\frac{90}{0.25t+1}\\But \ t=16\\\therefore X(t)=90-\frac{90}{0.025\times16+1}\\X(t)=25.71

5 0
2 years ago
2. Find the general relation of the equation cos3A+cos5A=0
mars1129 [50]
<h2>Answer:</h2>

A=\frac{\pi}{8}+\frac{n\pi}{4}or\ A=\frac{\pi}{2}+n\pi

<h2>Step-by-step explanation:</h2>

<h3>Find angles</h3>

cos3A+cos5A=0

________________________________________________________

<h3>Transform the expression using the sum-to-product formula</h3>

2cos(\frac{3A+5A}{2})cos(\frac{3A-5A}{2})=0

________________________________________________________

<h3>Combine like terms</h3>

2cos(\frac{8A}{2})cos(\frac{3A-5A}{2})=0\\\\  2cos(\frac{8A}{2})cos(\frac{-2A}{2})=0

________________________________________________________

<h3>Divide both sides of the equation by the coefficient of variable</h3>

cos(\frac{8A}{2})cos(\frac{-2A}{2})=0

________________________________________________________

<h3>Apply zero product property that at least one factor is zero</h3>

cos(\frac{8A}{2})=0\ or\ cos(\frac{-2A}{2})=0

________________________________________________________

<h2>Cos (8A/2) = 0:</h2>

<h3>Cross out the common factor</h3>

cos\ 4A=0

________________________________________________________

<h3>Solve the trigonometric equation to find a particular solution</h3>

4A=\frac{\pi}{2}or\ 4A=\frac{3\pi}{2}

________________________________________________________

<h3>Solve the trigonometric equation to find a general solution</h3>

4A=\frac{\pi}{2}+2n\pi \ or\\ \\ 4A=\frac{3 \pi}{2}+2n \pi\\ \\A=\frac{\pi}{8}+\frac{n \pi}{4\\}

________________________________________________________

<h2>cos(-2A/2) = 0</h2>

<h3>Reduce the fraction</h3>

cos(-A)=0

________________________________________________________

<h3>Simplify the expression using the symmetry of trigonometric function</h3>

cosA=0

________________________________________________________

<h3>Solve the trigonometric equation to find a particular solution</h3>

A=\frac{\pi }{2}\ or\ A=\frac{3 \pi}{2}

________________________________________________________

<h3>Solve the trigonometric equation to find a general solution</h3>

A=\frac{\pi}{2}+2n\pi\ or\ A=\frac{3\pi}{2}+2n\pi,n\in\ Z

________________________________________________________

<h3>Find the union of solution sets</h3>

A=\frac{\pi}{2}+n\pi

________________________________________________________

<h2>A = π/8 + nπ/4 or A = π/2 + nπ, n ∈ Z</h2>

<h3>Find the union of solution sets</h3>

A=\frac{\pi}{8}+\frac{n\pi}{4}\ or\ A=\frac{\pi}{2}+n\pi ,n\in Z

<em>I hope this helps you</em>

<em>:)</em>

5 0
2 years ago
A set of weights includes a 4 lb barbell and 6 pairs of weight plates. Each pair of plates weighs 20 lb. If x pairs of plates ar
Sunny_sXe [5.5K]

Answer:

Range[4,24,44,64,84,104,124]

Step-by-step explanation:

CHECK THE COMPLETE QUESTION BELOW;

A set of weights includes a 4 lb barbell and 6 pairs of weight plates. Each pair of plates weighs

20 lb. If x pairs of plates are added to the barbell, the total weight of the barbell and plates in

pounds can be represented by f x( ) = 20x + 4.

What is the range of the function for this situation?

The range of a function can be defined as the difference between the maximum and minimum function. From the question, we can deduced that the minimum function of X is 0 and no plate is added, and the maximum is 6

The given function is

f(x) = 20x + 4 =

f(0) = 20(0) + 4

= 4

f(6) = 20(6) + 4

= 124

Which means the range is within 4 and 124.

But we are told from the question that Each pair of plates weighs 20 lb, which means there is difference of 20ib each between 4 and 24

Then we can compute the range as [4,124] or (4,24,44,64,84,104,124)

3 0
2 years ago
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