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katovenus [111]
3 years ago
15

Help if possible, thanks.

Mathematics
1 answer:
lukranit [14]3 years ago
4 0

Answer:

1,2,3,6 are congruent

Step-by-step explanation:

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If the function h is defined by h(x)=<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^{2}" a
harkovskaia [24]

Given:

The function is:

h(x)=x^2-3x+5

To find:

The value of h(2x+1).

Solution:

We have,

h(x)=x^2-3x+5

Putting x=2x+1, we get

h(2x+1)=(2x+1)^2-3(2x+1)+5

h(2x+1)=(2x)^2+2(2x)(1)+(1)^2-3(2x)-3(1)+5

h(2x+1)=4x^2+4x+1-6x-3+5

On combining like terms, we get

h(2x+1)=4x^2+(4x-6x)+(1-3+5)

h(2x+1)=4x^2-2x+3

Therefore, the required function is h(2x+1)=4x^2-2x+3.

3 0
3 years ago
Write a word phrase to represent the numerical expression below. 5+ (17-8)
slavikrds [6]

Jimmy asked me for 8 of my 17 cookies so i gave them to him which left me with nine so I went to the store and got 5 which gives me 14

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7 0
3 years ago
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A tank contains 300 liters of fluid in which 40 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu
bonufazy [111]

Answer:

A(t) = 300 -260e^(-t/50)

Step-by-step explanation:

The rate of change of A(t) is ...

A'(t) = 6 -6/300·A(t)

Rewriting, we have ...

A'(t) +(1/50)A(t) = 6

This has solution ...

A(t) = p + qe^-(t/50)

We need to find the values of p and q. Using the differential equation, we ahve ...

A'(t) = -q/50e^-(t/50) = 6 - (p +qe^-(t/50))/50

0 = 6 -p/50

p = 300

From the initial condition, ...

A(0) = 300 +q = 40

q = -260

So, the complete solution is ...

A(t) = 300 -260e^(-t/50)

___

The salt in the tank increases in exponentially decaying fashion from 40 grams to 300 grams with a time constant of 50 minutes.

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3 years ago
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What is the gradient and y-intercept of a<br> line with equation y = 3x+2?
Diano4ka-milaya [45]
Gradient:3
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2 years ago
How many solutions douse -5x-8=+8
larisa [96]

Answer: One solution

Step-by-step explanation:

Given

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Divide -5 on both sides

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Therefore, it has one solution

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8 0
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