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3241004551 [841]
3 years ago
11

Label the figure using the drawing tools. Type the equation below.

Mathematics
1 answer:
Masteriza [31]3 years ago
8 0
The answer would be
4
5
6
Yes
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Order of operations and cross<br> reduce:<br> I got 20/3 but I think I’m wrong
poizon [28]

Answer:

130/81

simplified: 1 49/81

Step-by-step explanation:

I did the math

8 0
3 years ago
What are the possible numbers of positive, negative, and complex zeros of f(x) = −3x4 + 5x3 − x2 + 8x + 4?
ale4655 [162]

Answer:

Either:  1 neg, 3 pos, 0 imaginary; 1 neg, 1 pos, 2 imaginary

Step-by-step explanation:

Look for the positive possibilities first.  Count the numbe of sign changes then subtract 2, if possible, as many times as you can.

There are 3 sign changes.  So the possible positive roots are either 3 or 1.

Now look for the negative possibilities.  Replace each x with a -x and then count the sign changes.  Replacing with -x's gives you this polynomial:

f(-x)=-3x^4-5x^3-x^2-8x+4

There is only one sign change here, so the possible negative roots is 1.  Start with the negative roots to find the possible combinations of positive, negative, and imaginary, since there is only 1.

-     1       1

+    3       1

i     0       2

Since this is a 4th degree, the number of roots we have has to add up to equal 4.

8 0
3 years ago
Order each set of values from Least to Greatest Here is both Sets.
irina1246 [14]

Answer: mhhj

Step-by-step explanation:

Jdndndn

3 0
3 years ago
Read 2 more answers
Determine whether each expression is equivalent to 49^2t – 0.5.
vampirchik [111]

Answer:

None of the expression are equivalent to 49^{(2t - 0.5)}

Step-by-step explanation:

Given

49^{(2t - 0.5)}

Required

Find its equivalents

We start by expanding the given expression

49^{(2t - 0.5)}

Expand 49

(7^2)^{(2t - 0.5)}

7^2^{(2t - 0.5)}

Using laws of indices: (a^m)^n = a^{mn}

7^{(2*2t - 2*0.5)}

7^{(4t - 1)}

This implies that; each of the following options A,B and C must be equivalent to 49^{(2t - 0.5)} or alternatively, 7^{(4t - 1)}

A. \frac{7^{2t}}{49^{0.5}}

Using law of indices which states;

a^{mn} = (a^m)^n

Applying this law to the numerator; we have

\frac{(7^{2})^{t}}{49^{0.5}}

Expand expression in bracket

\frac{(7 * 7)^{t}}{49^{0.5}}

\frac{49^{t}}{49^{0.5}}

Also; Using law of indices which states;

\frac{a^{m}}{a^n} = a^{m-n}

\frac{49^{t}}{49^{0.5}} becomes

49^{t-0.5}}

This is not equivalent to 49^{(2t - 0.5)}

B. \frac{49^{2t}}{7^{0.5}}

Expand numerator

\frac{(7*7)^{2t}}{7^{0.5}}

\frac{(7^2)^{2t}}{7^{0.5}}

Using law of indices which states;

(a^m)^n = a^{mn}

Applying this law to the numerator; we have

\frac{7^{2*2t}}{7^{0.5}}

\frac{7^{4t}}{7^{0.5}}

Also; Using law of indices which states;

\frac{a^{m}}{a^n} = a^{m-n}

\frac{7^{4t}}{7^{0.5}} = 7^{4t - 0.5}

This is also not equivalent to 49^{(2t - 0.5)}

C. 7^{2t}\ *\ 49^{0.5}

7^{2t}\ *\ (7^2)^{0.5}

7^{2t}\ *\ 7^{2*0.5}

7^{2t}\ *\ 7^{1}

Using law of indices which states;

a^m*a^n = a^{m+n}

7^{2t+ 1}

This is also not equivalent to 49^{(2t - 0.5)}

6 0
3 years ago
Five members of the soccer team and five members of the track team ran the 100-meter dash. Their times are listed in the table b
viva [34]
The last one....0.74
5 0
3 years ago
Read 2 more answers
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