All categories

3 years ago

Find all real solutions of the equation. x4/3 − 7x2/3 + 10 = 0

1 answer:

8 0

Solve by Factoring x^(2/3)-7x^(1/3)+10=0
x
2
3
−
7
x
1
3
+
10
=
0
x
2
3
-
7
x
1
3
+
10
=
0
Rewrite
x
2
3
x
2
3
as
(
x
1
3
)
2
(
x
1
3
)
2
.
(
x
1
3
)
2
−
7
x
1
3
+
10
=
0
(
x
1
3
)
2
-
7
x
1
3
+
10
=
0
Let
u
=
x
1
3
u
=
x
1
3
. Substitute
u
u
for all occurrences of
x
1
3
x
1
3
.
u
2
−
7
u
+
10
=
0

You might be interested in

Log10 (2x + 1) - logio(3x - 2) = 1

Pavlova-9 [17]

Answer: x=3/4

Step-by-step explanation:

To solve, first express the left side as one logarithm. To do so, apply the quotient rule which is log_bM/N=log_bM-log_bN .

log((2x+1)/(3x-2))=1

Then, express the equation to its equivalent exponential form to eliminate the logarithm.

Note that the exponential form of log_b M=a is M=b^a .

Since the base of the logarithm in the given equation is not written, it indicates that its base is 10.

So re-writing the equation, it becomes:

log_10((2x+1)/(3x-2))=1

And its exponential form is:

(2x+1)/(3x-2)=10^1

(2x+1)/(3x-2)=10

Now that the equation has no more logarithm, the next step is to remove the x in the denominator.

To do so, multiply both sides by 3x-2.

(3x-2)*(2x+1)/(3x-2)=10*(3x-2)

2x+1=30x-20

Next, combine like terms.

To combine 30x and 2x, bring them together on one side of the equation. So, move 2x to the right side by subtracting both sides by 2x.

2x-2x+1=30x-2x-20

1=28x-20

To combine 20 and 1, bring them together on the side opposite the term with x. So, add both sides by 20.

1+20=28x-20+20

21=28x

And, divide both sides by 28 to have x only at the right side.

21/28=(28x)/28

3/4=x

Hence, the solution to the given equation is x=3/4 .

3 0

2 years ago

Evaluate a. 6! b. 8P5 c. 12C4

erik [133]

I'm pretty sure that the answer is 12C4 = (12p4)/4! = (12*11*10*9)/4*3*2*1 = 495

Hope this helps

4 0

2 years ago

Tammy deposited $520 in the bank account that earns simple interest every year after 5 years she had earned $156 and interest if

aleksley [76]

Answer:

6%.

Step-by-step explanation:

We have been given that Tammy deposited $520 in the bank account that earns simple interest every year after 5 years she had earned $156.

To find the interest rate we will use simple interest formula.

$I=P*r*T$

I= Interest.

P= Principal amount.

r=Annual interest rate (in decimal form).

T= Time in years.

We have been given that I=156, T=5, P=520

Upon substituting our values in above formula we will get,

$156=520*r*5$

$156=2600r$

$r=\frac{156}{2600}$

$r=0.06$

Let us multiply 0.06 by 100 to convert annual interest rate in percentage.

$0.06*100=6 \text{ percent}$

Therefore, the annual interest rate was 6%.

6 0

3 years ago

If the squared difference of the zeroes of the quadratic polynomial x2+kx+30 is equal to 169 find the value of k and the zeroes

anyanavicka [17]

ANSWER

$x = 2 \: \: or \: \: x = 15$

$x = - 2 \: \: or \: \: x = - 15$

EXPLANATION

The given polynomial is

$f(x) = {x}^{2} + kx + 30$

where a=1,b=k, c=30

Let the zeroes of this polynomial be m and n.

Then the sum of roots is

$m + n = - \frac{b}{a} = -k$

and the product of roots is

$mn = \frac{c}{a} = 30$

The square difference of the zeroes is given by the expression.

$( {m - n})^{2} = {(m + n)}^{2} - 4mn$

From the question, this difference is 169.

This implies that:

$( { - k)}^{2} - 4(30) = 169$

${ k}^{2} -120= 169$

$k^{2} = 289$

$k= \pm \sqrt{289}$

$k= \pm17$

We substitute the values of k into the equation and solve for x.

$f(x) = {x}^{2} \pm17x + 30$

$f(x) = (x \pm2)(x \pm 15)$

The zeroes are given by;

$(x \pm2)(x \pm 15) = 0$

$x = \pm2 \: \: or \: \: x = \pm 15$

5 0

3 years ago

3 x 104 + 2.5 x 105 = A. 0.5 x 105 B. 2.8 x 105 C. 3.25 x 105 D. 5.5 x 109

WITCHER [35]

Answer:

574.5 A) 52.5 B) 294 C) 341.25 D) 599.5

Step-by-step explanation:

3*104+2.5*105=312+262.5=574.5

A) 0.5*105=52.5

B) 2.8*105=294

C) 3.25*105=341.25

D) 5.5*109=599.5

6 0

3 years ago