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3 years ago

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Mathematics

1 answer:

Morgarella [4.7K]3 years ago

4 0

Answer: C. 18 1/16

Step-by-step explanation:

Square area = s^2

s = 4 1/4

4 1/4 * 4 1/4 = 18 1/16

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Write a quadratic function, in standard form, that fits the set of points. Solve it as a system of three equations. (-4, 9), (0,

Tresset [83]

Answer:

$\boxed{y=2x^2+4x-7}$

Step-by-step explanation:

Let the quadratic function be

$y=ax^2+bx+c$

We substitute $(-4,9)$ into the equation to obtain;

$9=a(-4)^2+b(-4)+c$

$\Rightarrow 9=16a-4b+c---(1)$

We substitute $(0,-7)$ to obtain;

$-7=a(0)^2+b(0)^2+c$

$\Rightarrow c=-7---(2)$

We finally substitute $(1,-1)$ to obtain;

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

$\Rightarrow -1=a+b+c---(3)$

We put equation (2) into equation (1) to get;

$9=16a-4b-7$

$16a-4b=16$

$\Rightarrow 4a-b=4---(4)$

$\Rightarrow -1=a+b-7$

$\Rightarrow a+b=6---(5)$

We add equation (4) and (5) to get;

$4a+a=6+4$

$\Rightarrow 5a=10$

$\Rightarrow a=2$

We put $a=2$ into equation (5) to get;

$2+b=6$

$\Rightarrow b=6-2$

$\Rightarrow b=4$

The reqiured quadratic function is

$y=2x^2+4x-7$

7 0

2 years ago

Simplify 28/184 by reducing to lowest form<br> A) 1/46<br> B) 1/7<br> C) 14/92<br> D) 7/46

Elodia [21]

It’ll be D) 7/46 you can reduce it by dividing everything with 4

5 0

2 years ago

The polynomial of degree 4, P ( x ) , has a root of multiplicity 2 at x = 1 and roots of multiplicity 1 at x = 0 and x = − 2 . I

ICE Princess25 [194]

We want to find a polynomial given that we know its roots and a point on the graph.

We will find the polynomial:

p(x) = (183/280)*(x - 1)*(x - 1)*(x + 2)*x

We know that for a polynomial with roots {x₁, x₂, ..., xₙ} and a leading coefficient a, we can write the polynomial equation as:

p(x) = a*(x - x₁)*(x - x₂)...*(x - xₙ)

Here we know that the roots are:

x = 1 (two times)
x = 0
x = -2

Then the roots are: {1, 1, 0, -2}

We can write the polynomial as:

p(x) = a*(x - 1)*(x - 1)(x - 0)*(x - (-2))

p(x) = a*(x - 1)*(x - 1)*(x + 2)*x

We also know that this polynomial goes through the point (5, 336).

This means that:

p(5) = 336

Then we can solve:

336 = a*(5 - 1)*(5 - 1)*(5 + 2)*5

336 = a*(4)*(4)*(7)*5

336 = a*560

366/560 = a = 183/280

Then the polynomial is:

p(x) = (183/280)*(x - 1)*(x - 1)*(x + 2)*x

If you want to learn more, you can read:

brainly.com/question/11536910

5 0

2 years ago

If 2x3 – 4x2 + kx + 10 is divided by (x + 2), the remainder is 4. Find the value of k using remainder theorem. Please help :)

Vinvika [58]

The polynomial remainder theorem states that the remainder of the division of a polynomial $P(x)$ by $x-a$ is equal to $P(a)$ .

Therefore

$P(-2)=4\\2\cdot(-2)^3 - 4\cdot(-2)^2 + k\cdot(-2) + 10=4\\-16-16-2k=-6\\-2k=26\\k=-13$

8 0

2 years ago

What number divided by 2 3 4 5 6 has a remainder of 1 but when divided by 7 has no remainder?

Afina-wow [57]

301

We could start by finding the lowest common multiple of 2, 3, 4, 5, and 6, which is 60. Then, we can consider the next few multiples: 120, 180, 240, 300...

However, because we need a remainder of 1 when our number is divided by each of these numbers (2,3,4,5,6), we want to go one above each of these multiples. So we're talking about 61, 121, 181, 241, 301... Those are the numbers that will satisfy the "remainder of 1" part of the question.

Now, we need to find out which one satisfies the other part of the question, which just requires dividing each of these numbers by 7 to see which is divisible by 7 (in other words, which one gives us a remainder of zero when we divide by 7).

301 does it. 301/7 = 43. So 301 is a multiple of 7 and therefore will yield no remainder when divided by 7.

Hope this all makes sense.

3 0

3 years ago