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3 years ago

6

HCFof 506 and 1863

1 answer:

Bogdan [553]3 years ago

4 0

Answer:

The hcf of 506 and 1863 is 23

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Evaluate the expression. ( – 5+ – 9)× – 3÷ – 2

Ronch [10]

Answer:

$-21$

Step-by-step explanation:

$( - 5+ - 9) \times - 3 \div - 2$

Solve the brackets first.

$( - 14) \times - 3 \div - 2$

Division comes next.

$( - 14) \times - 3/-2$

Cancel the negative signs.

$- 14 \times 3/2$

Multiply the terms.

$-42/2$

$=-21$

3 0

4 years ago

Read 2 more answers

Can someone help please!!

Tpy6a [65]

Answer:

Step-by-step explanation:

3 0

4 years ago

Need help with # 12 for distance/ midpoints formula. And another problem for #1 distance, segment and radicals ? Please help

Alenkinab [10]

Step-by-step explanation:

#12

PQ = 6x + 25

QR = 16 - 3x

PR = 6x + 25 + 16 - 3x

PR = 3x + 41

#1

D = √(-4 - 5)² + (-7 - 2)²

D = √(-9)² + (-9)²

D = √81 + 81

D = √162

D = 9√2

3 0

3 years ago

Find the real zeros 4x²-16x + 2.5x

Lapatulllka [165]

I am going to assume 2.5x is really 25x (typo).

First formulate the equality

$4x^2-16x+25x=0 \\4x^2+9x=0 \\x(4x+9)=0 \\$

Now two results emerge.

$x_1=0, x_2=-\dfrac{9}{4}$ .

Hope this helps.

8 0

3 years ago

The Vertex of a parabola is at (8-1), and it's why intercept is negative 17, which function represents the parabola

suter [353]

Answer:

The function is equal to $y=-(1/4)(x-8)^{2}-1$

Step-by-step explanation:

we know that

The equation of a vertical parabola in vertex form is equal to

$y=a(x-h)^{2}+k$

where

a is a coefficient

(h,k) is the vertex

In this problem we have

(h,k)=(8,-1)

substitute

$y=a(x-8)^{2}-1$

<u><em>Find the value of a</em></u>

Remember that we have the y-intercept

The y-intercept is the point (0,-17)

substitute

x=0,y=-17

$-17=a(0-8)^{2}-1$

$-17=64a-1$

$64a=-17+1$

$64a=-16$

$a=-16/64$

$a=-1/4$

therefore

The function is equal to

$y=-(1/4)(x-8)^{2}-1$

see the attached figure to better understand the problem

4 0

4 years ago