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3 years ago

Can someone help me with this please

Mathematics

1 answer:

Vinvika [58]3 years ago

3 0

Answer:

I': (-1, 4)

J': (3, 5)

K': (3, 2)

Step-by-step explanation:

to find the new 'x' values, add 2 to original 'x' values

to find the new 'y' values, add 6 to original 'y' values

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HELP PLZ PLZ PLZ!!! Maths

TiliK225 [7]

these inequalities can be solved just like equations

a. x - 7 > 25

+ 7 +7 x > 32the number line would have a filled in circle at 32 with an arrow pointing to the rightb. x - 8 < 5+8 +8x < 13 the number line would have an empty circle at 13 with an arrow pointing to the leftc.x + 11 > 23 - 11 - 11x > 12the number line would have an empty circle at 12 with an arrow pointing to the rightd. x + 12 < 6 - 12 - 12 x < -6the number line would have a filled in circle at -6 with an arrow pointing to the lefta. 21 < x + 18 -18 -18 3 < xx > 3b. 16 < x - 7 + 7 + 7 23 < x x > 23c. 12 > x - 7 + 7 + 7 19 > x x < 19d. 29 < x + 53 - 53 - 53 -24 < x x > -24e. 15 > x + 19 -19 -19 -4 > x x < -4f. 35 > x - 8 +8 +8 43 > x x < 43g. -6 < x - 18 +18 +18 12 < x x > 12h. -15 < x - 2 + 2 +2 -13 < x x > -13

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3 years ago

HURRY I'LL PICK A BRAINLIEST PERSON!!!

Rzqust [24]

Answer:

The only options which are having same solution as the given equation is A, B and D.

Explanation:

Given the equation: $\frac{3}{5}x+ \frac{2}{3} +x =\frac{1}{2}- \frac{1}{5}x$ ......[1]

Like terms are terms whose variables are the same.

In the given equation [1],

Combine like terms of variable x on LHS; we have

$\frac{3x+5x}{5}+ \frac{2}{3} =\frac{1}{2}- \frac{1}{5}x$

Simplify:

$\frac{8x}{5}+\frac{2}{3} =\frac{1}{2}- \frac{1}{5}x$

or we can write this as;

$\frac{8}{5}x +\frac{2}{3} =\frac{1}{2}- \frac{1}{5}x$ .

LCM(Least Common Multiple) to change each fraction to make their denominators the same as the least common denominator.

Taking LCM to both sides of an equation [1] as;

LCM of 3, 5 on LHS is 15 and LCM of 2 , 5 on RHS is 10;

$\frac{9x+10+15x}{15} = \frac{5-2x}{10}$

By cross multiplication we get;

$10 \cdot (9x+10+15x) = 15 \cdot (5-2x)$

Divide both side by 5 we get;

$2 \cdot (9x+10+15x) = 3 \cdot (5-2x)$

Using distributive property on both sides of an equation (i.e, $a\cdot (b+c) =a\cdot b+a\cdot c$ )

we have;

$18x+20+30x = 15-6x$ ......[2]

Additive Property of Equality states that allows one to add the same quantity to both sides of an equation.

Using additive property of equality:

Add 6x to both sides of an equation in [2];

$18x+20+30x+6x = 15-6x+6x$

Simplify;

$18x+20+30x+6x = 15$

Combine 18x and 6x we get;

$24x+20+30x= 15$ ......[3]

Subtraction Property of Equality states that allows one to subtract the same quantity to both sides of an equation.

Subtract 20 from both sides of an equation in [3] we get;

$24x+20+30x-20= 15-20$

Simplify:

$24x+30x= -5$

Therefore, from the given options only A, B and D have the same solution as $\frac{3}{5}x+ \frac{2}{3} +x =\frac{1}{2}- \frac{1}{5}x$

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3 years ago

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neonofarm [45]

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Step-by-step explanation:

(x⁴ + x)^½

As x approaches infinity, x⁴ becomes much larger than x. So:

(x⁴ + x)^½ ≈ (x⁴)^½ = x²

So x² grows at the same rate as (x⁴ + x)^½.

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The system of equations that represent the matrix is c

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Can someone help me with this please​

Can someone help me with this please