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3 years ago

Find the square root of 403225 by long division method.

Mathematics

1 answer:

Juli2301 [7.4K]3 years ago

5 0

Answer:

635 is the answer hope this help please give me Brainliest anwer

Step-by-step explanation:

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ASAP NEED HELP!!!!

alexira [117]

A. X+Y=50 y=x+20

B. 50-x=x+20 x=15 minutes

C. No, We know however much time Ted spends on the butterfly he spends 20 mins more on breast stroke. If he spends 45 mins on butterfly then he must spend 45+20= 65 mins on breast stroke. BUT we already know that he only spends 50 mins total, so it’s not possible

4 0

3 years ago

1-2 sin e cos 20-2 sin 40 = cos (20)

natka813 [3]

Answer:

No this is not the same as cos(2Ф)

Step-by-step explanation:

First I changed the $cos^{2}$ Ф into 1- $sin^{2}$ Ф

I then foiled the (1-2 $sin^{2}$ Ф)(1- $sin^{2}$ Ф) and that resulted in 1- $sin^{2}$ Ф-2 $sin^{2}$ Ф+2 $sin^{4}$ Ф

From there, I then cancelled the two 2 $sin^{4}$ Ф and combined like terms and was left with 1-3 $sin^{2}$ Ф

Now 1-3 $sin^{2}$ Ф ≠ cos(2Ф)

8 0

4 years ago

Let f(×)=2× and g(×)=2^× Graph the fuctions on the same coordinate plane. What are the solutions to the equation f(×)=g(×) Enter

kari74 [83]

Answer:

x=2, or x=1

Step-by-step explanation:

Given the equations as

f(x)=2x and

g(x)=2^x

Using a graph tool the equations will intersect at points (2,4) and (1,2)

x =2 or x=1

7 0

3 years ago

Evaluate the expression begin expression . . . the quantity, negative b plus b . . . times the quantity, 2 plus 7 . . . end expr

Kryger [21]

$- b + b \times (2 + 7) \\ = - 5 + 5 \times 9 \\ = - 5 + 45 \\ = 40$
The answer is 40

Hope this helps. - M

3 0

3 years ago

Find the inverse of the function. f(x) = the cube root of quantity x divided by seven. - 9

elena-14-01-66 [18.8K]

To solve

replace f(x) with y
switch x and y
solve for y
replace y with f⁻¹(x)

so
I'm not sure if it is
$f(x)=\sqrt[3]{\frac{x}{7}}-9$ or
$f(x)=\sqrt[3]{\frac{x}{7}-9}$

first one
$f(x)=\sqrt[3]{\frac{x}{7}}-9$
replace f(x) with y
$y=\sqrt[3]{\frac{x}{7}}-9$
switch x and y
$x=\sqrt[3]{\frac{y}{7}}-9$
solve for y
$x+9=\sqrt[3]{\frac{y}{7}}$
$(x+9)^3=\frac{y}{7}$
$7(x+9)^3=y$
replace y with f⁻¹(x)
$f^{-1}(x)=7(x+9)^3$

2nd one
$f(x)=\sqrt[3]{\frac{x}{7}-9}$
replce f(x) with y
$y=\sqrt[3]{\frac{x}{7}-9}$
switch x and y
$x=\sqrt[3]{\frac{y}{7}-9}$
solve for y
$x^3=\frac{y}{7}-9$
$x^3+9=\frac{y}{7}$
$7(x^3+9)=y$
replace y with f⁻¹(x)
$f^{-1}(x)=7(x^3+9)$

if you meant $f(x)=\sqrt[3]{\frac{x}{7}}-9$ then the inverse is $f^{-1}(x)=7(x+9)^3$

if you meant $f(x)=\sqrt[3]{\frac{x}{7}-9}$ then the inverse is $f^{-1}(x)=7(x^3+9)$

7 0

3 years ago