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Studentka2010 [4]
3 years ago
13

Help please!!!!!

Mathematics
1 answer:
Sergio [31]3 years ago
7 0

Answer:

  • a. image 1

Step-by-step explanation:

<u>Given function:</u>

  • f(x) = 2*3ˣ

<u>Plot two points and check which graph has those:</u>

  • f(0) = 2*3⁰ = 2

and

  • f(1) = 2*3¹ = 6

Points (0, 2) and (1, 6) are correct on the image 1 with blue graph

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Solve the following system of equations.<br> 6x -5y=13 &amp; 9y-15+2x=0
julsineya [31]
  • <em>Answer:</em>

<em>x = 3</em>

<em>y = 1</em>

  • <em>Step-by-step explanation:</em>

<em>Hi ! </em>

<em>6x - 5y = 13</em>

<em>9y - 15 + 2x = 0</em>

<em />

<em>6x - 5y = 13</em>

<em>2x + 9y = 15 | ×(-3)</em>

<em />

<em>6x - 5y = 13</em>

<em>- 6x - 27y = - 45</em>

<em>add</em>

<em />

<em>6x - 6x - 5y - 27y = 13 - 45</em>

<em>- 32y = - 32 | ×(-)</em>

<em>32y = 32</em>

<em>y = 1</em>

<em>replace y = 1</em>

<em>6x - 5(1) = 13</em>

<em>6x - 5 = 13</em>

<em>6x = 13 + 5</em>

<em>6x = 18</em>

<em>x = 3</em>

<em>Good luck !</em>

<em />

5 0
4 years ago
Find the least common multiple of 8x³y6z, 6x5z², and<br> 10yzº.
kvv77 [185]

Answer:

The Least Common Multiple (LCM) of 8x^3y6z,6x5z^2, \text{ and } 10yz^0 \text{ is } 240 x^3yz^2

Step-by-step explanation:

<u>Definition of LCM</u>

The LCM of a, b , c is the smallest multiplier that is divisible by a, b and c

Here the three terms are :

8x^3y6z

6x5z^2

10yz^0=10y        since z^0 = 1

Factoring using prime factorization we get  8x^3y6z = 2.2.2.x^3.y.2.3.z³

= 2^4\cdot \:3\cdot \:x^3\cdot \:y\cdot \:z   (1)

Factoring 6x5z^2  we get

2\cdot \:3\cdot \:5\cdot \:x\cdot \:z^2  (2)

Factoring 10yz^0 we get

2 \;\cdot\; 5y                 (z^0 = 1)   (3)

The LCM is the multiple of each of the highest power in each factor

2^4\;.\;3\;.\;5\;.x^2\;.\;y\;.z^2 = 240 x^3yz^2

8 0
2 years ago
A field is 53 1/3 yards wide. What is the width of the field in feet?
Umnica [9.8K]

Answer: The  width of the field = 160 feet

Step-by-step explanation:

We are given that , A field is 53\dfrac{1}{3} yards wide.

That means , The width of the field = 53\dfrac{1}{3} yards

Since , the fraction is in mixed for , so first we convert this into improper fraction.

53\dfrac{1}{3}=\dfrac{3(53)+1}{3}=\dfrac{160}{3}

Therefore , the width of the field = \dfrac{160}{3} yards

Since 1 yard = 3 feet.

⇒ the width of the field = \dfrac{160}{3}\times3 feet

Therefore , the width of the field = 160 feet

5 0
3 years ago
What’s the answer to the equation 10=7-m
Elenna [48]

7 - m = 10

-m = 3

<em>m = -3</em>

Hope this helps! :)

6 0
4 years ago
Read 2 more answers
Find ∂w/∂s and ∂w/∂t using the appropriate Chain Rule.
Vesnalui [34]

Answer:

<h3>The value of \frac{\partial w}{\partial s} is e^{3t}-18e^{2s+t}</h3><h3>The value of \frac{\partial w}{\partial t} is 3e^{3t}-9e^{2s+t}</h3><h3>The partial derivative at s=-5 and t=10 is \frac{\partial w}{\partial s} is e^{30}-18</h3><h3>The partial derivative at s=-5 and t=10 is \frac{\partial w}{\partial t}=3(e^{30}-3)</h3>

Step-by-step explanation:

Given that the Function point are w=y^3-9x^2y

x=e^s, y=e^t and s = -5, t = 10

<h3>To find \frac{\partial w}{\partial s} and \frac{\partial w}{\partial t}using the appropriate Chain Rule : </h3>

w=y^3-9x^2y  

Substitute the values of x and y in the above equation we get

w=(e^t)^3-9(e^s)^2(e^t)

w=e^{3t}-9e^{2s}.e^t

<h3>Now  partially differentiating w with respect to s by using chain rule we have </h3>

\frac{\partial w}{\partial ∂s}=e^{3t}-9(e^{2s}).2(e^t)

=e^{3t}-18e^{2s}.(e^t)

=e^{3t}-18e^{2s+t}

<h3>Therefore the value of \frac{\partial w}{\partial s} is e^{3t}-18e^{2s+t}</h3>

w=e^{3t}-9e^{2s}.e^t

<h3>Now  partially differentiating w with respect to t by using chain rule we have </h3>

\frac{\partial w}{\partial t}=e^{3t}.(3)-9e^{2s}(e^t).(1)

=3e^{3t}-9e^{2s+t}

<h3>Therefore the value of \frac{\partial w}{\partial t} is 3e^{3t}-9e^{2s+t}</h3>

Now put s-5 and t=10 to evaluate each partial derivative at the given values of s and t :

\frac{\partial w}{\partial s}=e^{3t}-18e^{2s+t}

=e^{3(10}-18e^{2(-5)+10}

=e^{30}-18e^{-10+10}

=e^{30}-18e^0

=e^{30}-18

<h3>Therefore the partial derivative at s=-5 and t=10 is \frac{\partial w}{\partial s} is e^{30}-18</h3>

\frac{\partial w}{\partial t}=3e^{3t}-9e^{2s+t}

=3e^{3(10)}-9e^{2(-5)+10}

=3e^{30}-9e{-10+10}

=3e^{30}-9e{0}

=3e^{30}-9

\frac{\partial w}{\partial t}=3(e^{30}-3)

<h3>Therefore the partial derivative at s=-5 and t=10 is \frac{\partial w}{\partial t}=3(e^{30}-3)</h3>
6 0
3 years ago
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