PLZ HELLP ASAP !!!!!!!!!!!!!!!

The sum of two numbers is greater than either number ?

KonstantinChe [14]

N+n<8

is that what you are looking for?

4 0

2 years ago

Filipo is building a rectangular sandbox . The length of the sandbox is 1 foot longer than twice the width of the sandbox .the p

Alla [95]

Answer:

Part A : D.) $2w + 2(2w+ 1) = 29$

Part B : Length of the sandbox is 10 feet.

Step-by-step explanation:

Given,

Perimeter = 29 ft

We need to find the equation for the perimeter and also the length of the sandbox.

Solution,

Let the width of the sandbox be 'w'.

Now as per question said;

The length of the sandbox is 1 foot longer than twice the width of the sandbox.

So we can say that;

Length = $2w+1$

Now we know that the perimeter is equal to the sum of twice of length and width.

framing in equation form, we get;

Perimeter = $2(2w+1)+2w$

we have given the perimeter, so on substituting the value, we get;

$29=2(2w+1)+2w$

Hence The equation used to find the width is $29=2(2w+1)+2w$ .

Now we solve for 'w'.

Applying distributive property, we get;

$4w+2+2w=29\\\\6w+2=29$

Subtracting both side by '2' we get

$6w+2-2=29-2\\\\6w=27$

Dividing both side by 6 we get;

$\frac{6w}{6}=\frac{27}{6}\\\\w=4.5\ ft$

Width of the sandbox = 4.5 ft

Length of the sandbox = $2w+1=2\times4.5 +1 = 9+1=10\ ft$

Hence Length of the sandbox is 10 feet.

3 0

2 years ago

Cos ( α ) = √ 6/ 6 and sin ( β ) = √ 2/4 . Find tan ( α − β )

Zina [86]

Answer:

$\purple{ \bold{ \tan( \alpha - \beta ) = 1.00701798}}$

Step-by-step explanation:

$\cos( \alpha ) = \frac{ \sqrt{6} }{6} = \frac{1}{ \sqrt{6} } \\ \\ \therefore \: \sin( \alpha ) = \sqrt{1 - { \cos}^{2} ( \alpha ) } \\ \\ = \sqrt{1 - \bigg( {\frac{1}{ \sqrt{6} } \bigg )}^{2} } \\ \\ = \sqrt{1 - {\frac{1}{ {6} }}} \\ \\ = \sqrt{ {\frac{6 - 1}{ {6} }}} \\ \\ \red{\sin( \alpha ) = \sqrt{ { \frac{5}{ {6} }}} } \\ \\ \tan( \alpha ) = \frac{\sin( \alpha ) }{\cos( \alpha ) } = \sqrt{5} \\ \\ \sin( \beta ) = \frac{ \sqrt{2} }{4} \\ \\ \implies \: \cos( \beta ) = \sqrt{ \frac{7}{8} } \\ \\ \tan( \beta ) = \frac{\sin( \beta ) }{\cos( \beta ) } = \frac{1}{ \sqrt{7} } \\ \\ \tan( \alpha - \beta ) = \frac{ \tan \alpha - \tan \beta }{1 + \tan \alpha . \tan \beta} \\ \\ = \frac{ \sqrt{5} - \frac{1}{ \sqrt{7} } }{1 + \sqrt{5} . \frac{1}{ \sqrt{7} } } \\ \\ = \frac{ \sqrt{35} - 1 }{ \sqrt{7} + \sqrt{5} } \\ \\ \purple{ \bold{ \tan( \alpha - \beta ) = 1.00701798}}$

8 0

2 years ago

Simplify the expression 1+2*[(3+1)*5+3]

Lady bird [3.3K]

1 + 2 *[(3+1)*5+3]

= 1 + 2*[(4)*5+3]

= 1 + 2*[20+3]

= 1 + 2 * 23

= 1 + 46

= 47

Answer

c. 47

7 0

2 years ago

Read 2 more answers

Please help me now ! Thank you

jarptica [38.1K]

First we'll do two basic steps. Step 1 is to subtract 18 from both sides. After that, divide both sides by 2 to get x^2 all by itself. Let's do those two steps now

2x^2+18 = 10
2x^2+18-18 = 10-18 <<--- step 1
2x^2 = -8
(2x^2)/2 = -8/2 <<--- step 2
x^2 = -4

At this point, it should be fairly clear there are no solutions. How can we tell? By remembering that x^2 is never negative as long as x is real.

Using the rule that negative times negative is a positive value, it is impossible to square a real numbered value and get a negative result.

For example
2^2 = 2*2 = 4
8^2 = 8*8 = 64
(-10)^2 = (-10)*(-10) = 100
(-14)^2 = (-14)*(-14) = 196

No matter what value we pick, the result is positive. The only exception is that 0^2 = 0 is neither positive nor negative.

So x^2 = -4 has no real solutions. Taking the square root of both sides leads to

x^2 = -4
sqrt(x^2) = sqrt(-4)
|x| = sqrt(4)*sqrt(-1)
|x| = 2*i
x = 2i or x = -2i
which are complex non-real values

5 0

3 years ago