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3 years ago

Help me with that last 3 plzzzzz

Mathematics

2 answers:

Molodets [167]3 years ago

8 0

12) Answer: 1 and 5/8

13) 3 and 1/6=19/6=57/18
1 and 7/9=16/9=32/18
(57/18)-(32/18)= 25/18
Answer: 25/18 or 1 and 7/18

14) 10.73+20.00= 30.73
30.73-16.89=13.84
Answer: $13.84

goldenfox [79]3 years ago

7 0

1 3/8 is left because 12 - 10 5/8 is 1 3/8

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Simplify 3 - b (7b + 2) + 3b - (11 - b)

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Answer: -7b² + 2b - 8

Step-by-step explanation:

Given expression

3 - b (7b + 2) + 3b - (11 - b)

Expand parentheses and apply the distributive property if necessary

=3 - b · 7b - b · 2 + 3b - 11 + b

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=-7b² + (3b - 2b + b) + (3 - 11)

= $\boxed{-7b^2+2b-8}$

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Read 2 more answers

NO LINKS!!! What is the equation of these two graphs?

S_A_V [24]

Answer:

$\textsf{1.} \quad y=-\dfrac{1}{30}x^2+\dfrac{1}{2}x+\dfrac{68}{15}\:\:\textsf{ where}\:\:x \geq \dfrac{15-\sqrt{769}}{2}\\\\\quad \textsf{or} \quad y=2\sqrt{x+5}$

$\textsf{2.} \quad y=-|x+1|+5$

Step-by-step explanation:

<h3>Question 1</h3>

Method 1 - modelling as a quadratic with restricted domain

Assuming that the points given on the graph are points that the curve passes through, the curve can be modeled as a quadratic with a limited domain. Please note that as the x-intercept has not been defined on the graph, I am not including this in this first method.

Standard form of a quadratic equation:

$y=ax^2+bx+c$

Given points:

(-4, 2)
(-1, 4)
(4, 6)

Substitute the given points into the equation to create 3 equations:

Equation 1 (-4, 2)

$\implies a(-4)^2+b(-4)+c=2$

$\implies 16a-4b+c=2$

Equation 2 (-1, 4)

$\implies a(-1)^2+b(-1)+c=4$

$\implies a-b+c=4$

Equation 3 (4, 6)

$\implies a(4)^2+b(4)+c=6$

$\implies 16a+4b+c=6$

Subtract Equation 1 from Equation 3 to eliminate variables a and c:

$\implies (16a+4b+c)-(16a-4b+c)=6-2$

$\implies 8b=4$

$\implies b=\dfrac{4}{8}$

$\implies b=\dfrac{1}{2}$

Subtract Equation 2 from Equation 3 to eliminate variable c:

$\implies (16a+4b+c)-(a-b+c)=6-4$

$\implies 15a+5b=2$

$\implies 15a=2-5b$

$\implies a=\dfrac{2-5b}{15}$

Substitute found value of b into the expression for a and solve for a:

$\implies a=\dfrac{2-5(\frac{1}{2})}{15}$

$\implies a=-\dfrac{1}{30}$

Substitute found values of a and b into Equation 2 and solve for c:

$\implies a-b+c=4$

$\implies -\dfrac{1}{30}-\dfrac{1}{2}+c=4$

$\implies c=\dfrac{68}{15}$

Therefore, the equation of the graph is:

$y=-\dfrac{1}{30}x^2+\dfrac{1}{2}x+\dfrac{68}{15}$

$\textsf{with the restricted domain}: \quad x \geq \dfrac{15-\sqrt{769}}{2}$

Method 2 - modelling as a square root function

Assuming that the points given on the graph are points that the curve passes through, and the x-intercept should be included, we can model this curve as a square root function.

Given points: