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Tanya [424]
2 years ago
10

Let the function f be defined by f(x)=12-5x b)evaluate f(-1)

Mathematics
2 answers:
Alexxx [7]2 years ago
6 0
F(-1) = 12 - 5(-1)
f(-1) = 12 + 5
Solution: f(-1) = 17
OverLord2011 [107]2 years ago
5 0
F(x) = 12-5(-1)
= 12-(-5)
= 17
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Solve x2 – 8x + 15 &lt; 0.<br><br> Select the critical points for the inequality shown.
quester [9]

Answer:

x=3,5

Explanation:

x2−8x+15=0

Try to express the terms of the equation in square form.

Adding 16 both sides of the equation,

(x2−2⋅x⋅4+42)+15=16

or,(x−4)2+15−16=0

or,(x−4)2−1=0

or,(x−4)2−12=0

This is the a2−b2=(a+b)(a−b)form.

(x−4+1)(x−4−1)=0

or,(x−3)(x−5)=0

Now, equate both the terms to zero since both of them when multiplied, give zero.

Either,

x−3=0

∴x=3

Or,

x−5=0

∴x=5

Ans:x=3,5 Hope this helpsXD...!!

4 0
2 years ago
Read 2 more answers
Q.6. The equation of the ellipse whose centre is at the origin and the x-axis, the major axis, which passes
azamat

<h3>Answer:</h3>

Equation of the ellipse = 3x² + 5y² = 32

<h3>Step-by-step explanation:</h3>

<h2>Given:</h2>

  • The centre of the ellipse is at the origin and the X axis is the major axis

  • It passes through the points (-3, 1) and (2, -2)

<h2>To Find:</h2>

  • The equation of the ellipse

<h2>Solution:</h2>

The equation of an ellipse is given by,

\sf \dfrac{x^2}{a^2} +\dfrac{y^2}{b^2} =1

Given that the ellipse passes through the point (-3, 1)

Hence,

\sf \dfrac{(-3)^2}{a^2} +\dfrac{1^2}{b^2} =1

Cross multiplying we get,

  • 9b² + a² = 1 ²× a²b²
  • a²b² = 9b² + a²

Multiply by 4 on both sides,

  • 4a²b² = 36b² + 4a²------(1)

Also by given the ellipse passes through the point (2, -2)

Substituting this,

\sf \dfrac{2^2}{a^2} +\dfrac{(-2)^2}{b^2} =1

Cross multiply,

  • 4b² + 4a² = 1 × a²b²
  • a²b² = 4b² + 4a²-------(2)

Subtracting equations 2 and 1,

  • 3a²b² = 32b²
  • 3a² = 32
  • a² = 32/3----(3)

Substituting in 2,

  • 32/3 × b² = 4b² + 4 × 32/3
  • 32/3 b² = 4b² + 128/3
  • 32/3 b² = (12b² + 128)/3
  • 32b² = 12b² + 128
  • 20b² = 128
  • b² = 128/20 = 32/5

Substituting the values in the equation for ellipse,

\sf \dfrac{x^2}{32/3} +\dfrac{y^2}{32/5} =1

\sf \dfrac{3x^2}{32} +\dfrac{5y^2}{32} =1

Multiplying whole equation by 32 we get,

3x² + 5y² = 32

<h3>Hence equation of the ellipse is 3x² + 5y² = 32</h3>
8 0
3 years ago
Please answer the picture
dimaraw [331]
The area is equal to 230 CM.
area for rectangle :
area = width times length
area = 7 x 20 = 140CM
area for triangle:
area = 1/2 times base times height
area = 1/2 x 20 x 9 = 90CM

ADD:
140+90 = 230 CM
3 0
2 years ago
Read 2 more answers
Solve equation:<br><img src="https://tex.z-dn.net/?f=y%20%2B%20%20%5Cfrac%7B7%7D%7B8%7D%20%20%3D%20%20%5Cfrac%7B3%7D%7B8%7D%20"
Gala2k [10]

Answer: y = -1/2

Step-by-step explanation:

Subtract both sides by 7/8

y = 3/8 - 7/8

Then combine like terms

y = -4/8

Then simplify

y = -1/2

4 0
3 years ago
Help ASAP will give brainly for CORRECT ANSWER
Pie

Answer:

2cnd one 3rd one 4th one last one

Step-by-step explanation:

4 0
2 years ago
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