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Arte-miy333 [17]

3 years ago

11

Use doubles to find the sum of 7+8.

2 answers:

lyudmila [28]3 years ago

6 0

I believe, 8+8=16, 16-1=15

Tems11 [23]3 years ago

4 0

7+8 =15
so u can use 7+7 +1
since 15 is off number there is no exact double number for that u have to use 1

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Convert -18° to radians.

Fiesta28 [93]

0.314159

Is the answer

5 0

3 years ago

Find the length of side x in simplest radical form with a rational denominator.

kogti [31]

Answer:

since it is Right angled isosceles triangle it's base side are equal

by using Pythagoras law

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6 0

3 years ago

HELP ME ASAP WITH 8! PLS!! I beg u..

lakkis [162]

7. x=3 is the midpoint between the roots. The other root is x = 2*3 -(-5) = 11.

8a) f(x) = (x +3)^2 -49. The vertex is (-3, -49). The roots are -10, 4.

8b) y = (x+4)^2 -1. The vertex is (-4, -1). The roots are -5, -3.

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7 0

3 years ago

Which expression is equivalent to

lutik1710 [3]

Answer:

$\frac{\sqrt[4]{3x^2} }{2y}$

Step-by-step explanation:

We can simplify the expression under the root first.

Remember to use $\frac{a^x}{a^y}=a^{x-y}$

Thus, we have:

$\sqrt[4]{\frac{24x^{6}y}{128x^{4}y^{5}}} \\=\sqrt[4]{\frac{3x^{2}}{16y^{4}}}$

We know 4th root can be written as "to the power 1/4th". Then we can use the property $(ab)^{x}=a^x b^x$

<em>So we have:</em>

<em> $\sqrt[4]{\frac{3x^{2}}{16y^{4}}} \\=(\frac{3x^{2}}{16y^{4}})^{\frac{1}{4}}\\=\frac{3^{\frac{1}{4}}x^{\frac{1}{2}}}{2y}\\=\frac{\sqrt[4]{3x^2} }{2y}$ </em>

<em />

<em>Option D is right.</em>

8 0

4 years ago

What are the real and complex solutions of the polynomial equation? x^4-41x^2=-400. *work needed*

pickupchik [31]

<h3>Answer: x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>

Step-by-step explanation:

Given the polynomial equation $x^4-41x^2=-400$ .

Adding 400 on both sides to get rid 400 from right side and set 0 on right side, we get

$x^4-41x^2+400=-400+400$ .

$x^4-41x^2+400=0$ .

Factoring by product sum rule.

We need product of 400 and sum upto -41.

We can see that 400 = -25 × -16 = 400 and -25-16 = -41.

Therefore,

$x^4-25x^2-16x^2+400=0$

Making it into two groups, we get

$(x^4-25x^2)+(-16x^2+400)=0$

Factoring out GCF of each group, we get

$x^2(x^2-25)-16(x^2-25)=0$

$(x^2-25)(x^2-16) =0$

Factoring out $(x^2-25)$ and $(x^2-16)$ separately by difference of the squares identity $a^2-b^2=(a-b)(a+b)$ , we get

$(x^2-25) = x^2-5^2= (x-5)(x+5)$ and

$x^2-16 = x^2-4^2 =(x-4)(x+4)$ .

Therefore,

$(x-5)(x+5)(x-4)(x+4) =0$

Applying zero product rule,

x-5=0

x+5=0

x-4=0 and

x+4=0.

Therefore,

<h3>x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>

6 0

3 years ago