Find the missing base.

Mathematics

2 answers:

Leto [7]3 years ago

8 0

The answer is 2. 2^6=64

Svetllana [295]3 years ago

3 0

Let base be x

x^6 = 64

64 = 2*2*2*2*2*2

so the base x is 2

You might be interested in

Carlos and Diya perform the same transformations on

KengaRu [80]

Answer: Carlos is right

Step-by-step explanation:

why because in dilation, the image and the original are similar, in that they are the same shape but not necessarily the same size. They are not congruent because that requires them to be the same shape and the same size, which they are not (unless the scale factor happens to be 1.0).

Diya is wrong because the picture is the not the same size && the image is going to stay the same

8 0

3 years ago

(60 points)

natka813 [3]

Answer:

Step-by-step explanation:

$2.\\(a-b)^2=a^2-2ab+b^2\\\\x^2-16x=-8\\\\x^2-2(x)(8)=-8\\\\\text{We have}\ 2ab=2(x)(8).\ \text{Therefore}\ b=8.\\\\x^2-2(x)(8)=-8\qquad\text{add}\ 8^2=64\ \text{to both sides}\\\\x^2-2(x)(8)+8^2=-8+64\\\\(x-8)^2=56$

$5.\\2x^2+x-1=2\qquad\text{subtract 2 from both sides}\\\\2x^2+x-3=0\\\\2x^2+3x-2x-3=0\\\\x(2x+3)-1(2x+3)=0\\\\(2x+3)(x-1)=0\iff 2x+3=0\ \vee\ x-1=0\\\\2x+3=0\qquad\text{subtract 3 from both sides}\\2x=-3\qquad\text{divide both sides by 2}\\\boxed{x=-\dfrac{3}{2}}\\\\x-1=0\qquad\text{add 1 to both sides}\\\boxed{x=1}$

$6.\\2x^2-4x=0\qquad\text{divide both sides by 2}\\\\x^2-2x=0\\\\x(x-2)=0\iff x=0\ \vee\ x-2=0\\\\\boxed{x=0}\\\\x-2=0\qquad\text{add 2 to both sides}\\\boxed{x=2}$