All categories

3 years ago

Evaluate 3a2 + 4b2 for a = 2 and b = 3

Mathematics

1 answer:

Crank3 years ago

4 0

3a^2 + 4b^2 for a = 2 and b = 3

3(2)^2 + 4(3)^2 = 12 + 36 = 48

You might be interested in

Please help I need help

JulijaS [17]

The rate of change in this pattern is 2:1 comparing the change in 'y' to the change in 'x'.. this means that if you go form 33 to 51 ( an increase of 18), then the the increase from 16 to ? will be 9. Therefore, ? = 16 + 9 or 25
? = 25

4 0

2 years ago

Franklin bought a book with 90 pages. If he

BartSMP [9]

The answer is A) 60%

3 0

2 years ago

Diya spent 2/5 of her money on a dress and 1/2 of the reminder on a doll. She spent $8 more o the dress than the doll. How much

Elena L [17]

Answer:

$24

Step-by-step explanation:

2/5 — dress

3/5 — remainder

1/2 of remainder = 1/2 × 3/5 = 3/10 — doll

rewrite fraction spent on dress: 4/10

dress - doll = $8

4/10 - 3/10 = 1/10

1/10 = $8

fraction of money left = 10/10 - 4/10 - 3/10

= 3/10

amount of money left = $8 × 3

$24

8 0

2 years ago

What’s the vertical and horizontal line?

coldgirl [10]

Answer:

Part 1) Vertical line : $x=1$

Part 2) Horizontal line : $y=-4$

Step-by-step explanation:

Part 1) Write the equation for the vertical line passing through the point (1,-4)

we know that

The equation of a vertical line (parallel to the y-axis) is equal to the x-coordinate of the point that passes through it

The x-coordinate is 1

therefore

The equation of the line is

$x=1$

Part 2) Write the equation for the horizontal line passing through the point (1,-4)

we know that

The equation of a horizontal line (parallel to the x-axis) is equal to the y-coordinate of the point that passes through it

The y-coordinate is -4

therefore

The equation of the line is

$y=-4$

5 0

2 years ago

What are the zeros of the polynomial function: f(x) = x3 – 5x2 -6x

shepuryov [24]

The zeros of the polynomial function: f(x) = x^3 – 5x^2 -6x is -1, 0, 6.

Solution:

Given, polynomial equation is $x^{3}-5 x^{2}-6 x$

We have to find the zeroes of the given polynomial.

So, let us equate it with 0.

$\begin{array}{l}{\text { Then, } x^{3}-5 x^{2}-6 x=0} \\\\ {\rightarrow x\left(x^{2}-5 x-6\right)=0} \\\\ {\rightarrow x\left(x^{2}-(6-1) x-6\right)=0} \\\\ {\rightarrow x\left(x^{2}-6 x+x-6\right)=0} \\\\ {\rightarrow x(x(x-6)+1(x-6))=0} \\\\ {\rightarrow x(x-6)(x+1)=0} \\\\ {\rightarrow x=0 \text { or } x-6=0 \text { or } x+1=0} \\\\ {\rightarrow x=0 \text { or } 6 \text { or }-1}\end{array}$

Hence, the roots of the given polynomial are -1, 0, 6.

6 0

2 years ago