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3 years ago

What number is 4 times as many as 25

Mathematics

2 answers:

Sonbull [250]3 years ago

6 0

100. 25 4 times is 100. like 25 added 4 times

prohojiy [21]3 years ago

3 0

4 times as many as 25 is 100

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Which equation is represented by the graph below?

lesya692 [45]

Answer:

Step-by-step explanation:

The lines positioning shows a negative slope already, so it cant be C or D

Slope is a rise over run basis. You can check this just from points alone, the line would be much steeper if the slope was -2.

Hope this helps!

8 0

4 years ago

10. The graph of y=f(x) is given below.y!24168848 xWhat is the possible degree of f?A. 4IB. -3C. 2D. 3E. -1

steposvetlana [31]

Answer:

The degree of the function is;

$3$

Explanation:

From the given graph, we can observe that the function has two extremum (one minima and one maxima).

The degree of the function will be the number of extremum plus 1;

$\text{degree}=n+1$

Since there are two extremum on the graph then;

$\begin{gathered} \text{degree}=2+1 \\ \text{degree}=3 \end{gathered}$

Therefore, the degree of the function is;

$3$

4 0

1 year ago

Whats 1+12 fee ponitns

Dafna11 [192]

Answer:

Step-by-step explanation:

7 0

3 years ago

Find an equation of the line that satisfies the given conditions.

Makovka662 [10]

Well, the very first thing you need to do is to <span> find the slope of the line
According to the fact that we already know that the slope is parallel </span>to the line, you can easily find out the slope, to be more exact : $m=(Y2-Y1)/X2-X1)=4/4=1$
Straight after, you have to use <span>point slope formula</span> $Y-Y1=m(X-X1)

Y-7=X-6$

6 0

3 years ago

Recursive definitions for subsets of binary strings.Give a recursive definition for the specified subset of the binary strings.

dmitriy555 [2]

Answer:

Step-by-step explanation:

A binary string with 2n+1 number of zeros, then you can get a binary string with 2n(+1)+1 = 2n+3 number of zeros either by adding 2 zeros or 2 1's at any of the available 2n+2 positions. Way of making each of these two choices are (2n+2)22. So, basically if b2n+12n+1 is the number of binary string with 2n+1 zeros then your

b2n+32n+3 = 2 (2n+2)22 b2n+12n+1

your second case is basically the fact that if you have string of length n ending with zero than you can the string of length n+1 ending with zero by:

1. Either placing a 1 in available n places (because you can't place it at the end)

2. or by placing a zero in available n+1 places.

0 ϵ P

x ϵ P → 1x ϵ P , x1 ϵ P

x' ϵ P,x'' ϵ P → xx'x''ϵ P

3 0

3 years ago