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natka813 [3]
3 years ago
13

Find a and b given f(x) = ax + b, f(1) = 7, and f(3) = -5.

Mathematics
1 answer:
sergejj [24]3 years ago
3 0

Answer:

a = - 6, b = 13

Step-by-step explanation:

Given f(1) = 7 , that is f(x) = 7 when x = 1

f(3) = - 5 , that is f(x) = - 5 when x = 3

Substitute these values into f(x) = ax + b

a + b = 7 → (1)

3a + b = - 5 → (2)

Subtract (1) from (2) term by term to eliminate b

(3a - a) + (b - b) = - 5 - 7

2a = - 12 ( divide both sides by 2 )

a = - 6

Substitute a = - 6 into either of the 2 equations and solve for b

Substituting into (1)

- 6 + b = 7 ( add 6 to both sides )

b = 13

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Solve the following equation for a. Be sure to take into account whether a letter is
Zina [86]

Answer:

a = \frac{r-q}{g}

Step-by-step explanation:

Given

ga + q = r ( subtract q from both sides )

ga = r - q ( isolate a by dividing both sides by g )

a = \frac{r-q}{g}

5 0
2 years ago
The two lines y= 6x+15 and y= mx+4 intersect at x= -2. What is the y-coordinate of their intersection point
taurus [48]

Answer:

The y-coordinate of their intersection point is 3

That is y=3

Step-by-step explanation:

Given two lines are y=6x+15 and y=mx+4

Given that the two lines intersect at x=-2

To find the y coordinate of their intersection point :

Equating the two lines

6x+15=mx+4

6x+15-mx-4=0

6x-mx+11=0

(6-m)x+11=0

At x=-2  (6-m)x+11=0

(6-m)(-2)+11=0

(6-m)(-2)=-11

6-m=\frac{11}{2}

-m=\frac{11}{2}-6

-m=\frac{11-12}{2}

-m=\frac{-1}{2}

m=\frac{1}{2}

Substitute the value m=\frac{1}{2} in y=mx+4 we get

y=\frac{1}{2}x+4

At x=-2 y=\frac{1}{2}(-2)+4

=-1+4

=3

Therefore y=3

Therefore the y-coordinate of their intersection point is 3

7 0
3 years ago
Please help me answer this.
adoni [48]

Answer:

See explanation

Step-by-step explanation:

        Statements                         Reasons

1. \overline{AD}\parallel \overline{BC}                                        Given

2. \angle ADB\cong \angle CBD      As alternate interior angles when parallel lines AD and BC intersect by ltransversal BD

3. \overline{AD}\cong \overline{BC}                                     Given

4. \overline{BD}\cong \overline{DB}                     Reflexive property

5. \triangle ADB\cong \triangle CBD                  SAS postulate

6.  \angle ABD\cong \angle CDB              Corresponding parts of congruent triangles are congruent

7. \overline{AB}\parallel \overline{CD}            Inverse alternate interior angles theorem

8 0
3 years ago
Can someone please give me the answer to this problem?
Alik [6]
X= 9/2, 15/2
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3 0
2 years ago
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Answer:

a

Step-by-step explanation:

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