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3 years ago

Solve 3^4x+5=(1/3)^x+10 for x. please show work

Mathematics

1 answer:

loris [4]3 years ago

6 0

(3/4)/(1/3)=9/4
5-10=-5
(9/4)-5=x=-2.75

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Which value, when placed in the box, would result in a system of equations with infinitely many solutions?

ira [324]

<span>For a system of equations to have infinitely many solutions, the equations have to be lineary dependent, (i.e. one of the equations is a multiple of the other equation). For the given equation to have infinitely many solutions, the second equation have to be a multiple of the first equation. The first equation can be rewritten as y - 2x = -5 The second equation is the first equation multiplied by 2, i.e. 2(y - 2x= -5) = 2y - 4x = -10 Therefore, the correct answer is -10 (option a).</span>

7 0

3 years ago

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In the right triangle shown, the third side can be represented by the expression

Westkost [7]

The third side can be represented by D) $x = \sqrt{z^{2} } -y^{2} .$

Step-by-step explanation:

Step 1:

According to the Pythagorean theorem, the square of the hypotenuse will be equal to the sum of the squares of the other two sides.

In the given triangle, the hypotenuse is the side that is represented by z. Another side on the triangle is represented by the value y.

Step 2:

Assume the other side of the triangle measures x units.

So according to the Pythagorean theorem,

$z^{2} = x^{2} +y^{2} , x^{2} = z^{2} - y^{2} .$

$x = \sqrt{z^{2} } -y^{2}$ which is option D.

6 0

3 years ago

120° angle

quester [9]

Answer:

The equation has zero solutions because the equation 2 = 3 is never true.

Hope it helps :)

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2 years ago

How many integers between 10 and 60, inclusive, can be evenly divided by neither 2 nor 3?

erastovalidia [21]

Find total number of integers.

$a_1=10,a_n=60,d=1 \\a_n=a_1+(n-1)d \\60=10+(n-1)1 \\n-1=50 \\n=51$

Find how many integers is divisible by 2.

$a_1=10,a_n=60,d=2
\\a_n=a_1+(n-1)d
\\60=10+(n-1)2
\\2(n-1)=50
\\n-1=25
\\n=26$

Eliminate even numbers.

11, 13, 15,..., 57, 59

This array contains 51 - 26 = 25 numbers.

Eliminate numbers before the first number divisible by 3 and after the last number divisible by 3.

15, 17, 19,..., 55, 57

This array contains 25 - 3 = 22 numbers.

Now we should eliminate numbers divisible by 3: 15, 21, 27...

$a_1=15,a_n=57,d=6,n=?
\\a_n=a_1+(n-1)d
\\57=15+(n-1)6
\\6(n-1)=42
\\n-1=7
\\n=8$

There are 8 such numbers.

Therefore, there are 25 - 8 = 17 numbers that <span>can be evenly divided by neither 2 nor 3</span>

6 0

3 years ago

What is 8x8x8x8x3x3 in exponential form?

Morgarella [4.7K]

Step-by-step explanation:

$8 ^{4} \times 3 {}^{2}$

Count how many of the same number, since you have four 8's the exponent will be

$8 {}^{4}$

same applies for 3

4 0

3 years ago

Read 2 more answers