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3 years ago

Find the value of x for ABCD is a parallelogram

Mathematics

1 answer:

Ilya [14]3 years ago

8 0

Find parallelogram ABCD attached

Answer:

x = 3

Step-by-step explanation:

In parallelograms, the diagonals typically divide each other into 2 equal parts.

This means that from the attached parallelogram, we can say that;

4x - 6 = 2x or 6x - 15 = x

Thus;

4x - 2x = 6

2x = 6

x = 6/2

x = 3

Let's confirm with the second equation;

6x - x = 15

5x = 15

x = 15/5

x = 3

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Given that f(x) = x2 – 3x - 10 and g(x) = x – 5, find (f - g)(x) and express

Roman55 [17]

Answer:

$(f-g)(x)= x^2-4x-5$

Step-by-step explanation:

We have the two functions:

$f(x)=x^2-3x-10\text{ and } g(x)=x-5$

And we want to find (f-g)(x).

(f-g)(x) is equivalent to f(x)-g(x).

Therefore:

$\begin{aligned} (f-g)(x)&=f(x)-g(x)\\ &=(x^2-3x-10)-(x-5)\end{aligned}$

Simplify:

$\begin{aligned} &=x^2-3x-10-x+5 \\ &=(x^2)+(-3x-x)+(-10+5) \\ &=x^2-4x-5\end{aligned}$

Therefore:

$(f-g)(x)= x^2-4x-5$

8 0

3 years ago

Read 2 more answers

Factorise each of the following algebraic expressions completely,

LekaFEV [45]

Answer:

see explanation

Step-by-step explanation:

(a)

Given

2k - 6k² + 4k³ ← factor out 2k from each term

= 2k(1 - 3k + 2k²)

To factor the quadratic

Consider the factors of the product of the constant term ( 1) and the coefficient of the k² term (+ 2) which sum to give the coefficient of the k- term (- 3)

The factors are - 1 and - 2

Use these factors to split the k- term

1 - k - 2k + 2k² ( factor the first/second and third/fourth terms )

1(1 - k) - 2k(1 - k) ← factor out (1 - k) from each term

= (1 - k)(1 - 2k)

1 - 3k + 2k² = (1 - k)(1 - 2k) and

2k - 6k² + 4k³ = 2k(1 - k)(1 - 2k)

(b)

Given

2ax - 4ay + 3bx - 6by ( factor the first/second and third/fourth terms )

= 2a(x - 2y) + 3b(x - 2y) ← factor out (x - 2y) from each term

= (x - 2y)(2a + 3b)

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3 years ago

What is 25% of success as a fraction and a decimal

Tpy6a [65]

25/100 simplify and get 1/4
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4 years ago

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Deffense [45]

Answer:

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3 years ago

The value of y varies directly as the square of x and y=27 when x=9. Find the equation represents this relationship. Report the

skad [1K]

Answer:

The relation between x and y is

$y=\frac13 x^2$

Step-by-step explanation:

Given that,

The value of y varies directly as the square of x.

$y\propto x^2$

$\Rightarrow y=rx^2$

where r is constant of proportional.

Given that,

y= 27 and x=9

Plug y= 27 and x=9 in above relation

$y=rx^2$

$\therefore 27=r(9)^2$

$\Rightarrow 27 = r.81$

$\Rightarrow r=\frac{27}{81}$

$\Rightarrow r=\frac{1}{3}$

The relation between x and y is

$y=\frac13 x^2$

8 0

3 years ago