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Aleksandr [31]
3 years ago
9

When adding decimals with the same sign you.....​

Mathematics
2 answers:
PilotLPTM [1.2K]3 years ago
8 0

Answer:

keep the same sign and add the value to each number

Step-by-step explanation:

Stels [109]3 years ago
7 0

Answer:

keep the decemal in the same place? there's not really alot to go by on your question

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Use mental math to find the unknown number in the equation 12 x b= 480
makkiz [27]
So find out what times twelve equals 48. Then add a zero in front and that is your answer.
5 0
3 years ago
Read 2 more answers
B) Using the two point above find the slope using the formula m =
12345 [234]

The linear equations are y - 25 = 0.89(x - 20) and y  = 0.89x + 7.2

<h3>The slope of the line</h3>

The complete question is added as an attachment

The two points from the graph are (20, 25) and (38, 41)

The slope of the line is calculated using

m = (y2 - y1)/(x2 - x1)

Substitute the known values in the above equation

m = (41 - 25)/(38 - 20)

Evaluate

m =0.89

<h3>The linear equation in point slope form</h3>

This is calculated as:

y - y1 = m(x - x1)

Substitute the known values in the above equation

y - 25 = 0.89 * (x - 20)

Evaluate

y - 25 = 0.89(x - 20)

<h3>The linear equation in slope-intercept form</h3>

We have:

y - 25 = 0.89(x - 20)

Expand

y - 25 = 0.89x - 17.8

Add 25 to both sides

y  = 0.89x + 7.2

Hence, the linear equations are y - 25 = 0.89(x - 20) and y  = 0.89x + 7.2

Read more about linear equations at:

brainly.com/question/4025726

#SPJ1

5 0
2 years ago
Please help asap<br><br> solve x^4+5x^2+4=0
const2013 [10]

Answer:

x=\pm i\text{ or }x=\pm2i

Step-by-step explanation:

So we have the equation:

x^4+5x^2+4=0

Let's let u be equal to x². So:

u^2+5u+4=0

Factor:

(u+1)(u+4)=0

Zero Product Property:

u+1=0\text{ or } u+4=0

Subtract:

u=-1\text{ or }u=-4

Replace:

x^2=-1\text{ or }x^2=-4

Take square root:

x=\pm\sqrt{-1}\text{ or }x=\pm\sqrt{-4

Simplify:

x=\pm i\text{ or }x=\pm2i

7 0
3 years ago
Consider the sequence of numbers: 3/8, 3/4, 1 /18, 1 1/2, 1 7/8
vredina [299]

Question:

Consider the sequence of numbers: \frac{3}{8}, \frac{3}{4}, 1\frac{1}{8}, 1\frac{1}{2}, 1\frac{7}{8}

Which statement is a description of the sequence?

(A) The sequence is recursive, where each term is 1/4 greater than its preceding term.

(B) The sequence is recursive and can be represented by the function

f(n + 1) = f(n) + 3/8 .

(C) The sequence is arithmetic, where each pair of terms has a constant difference of 3/4 .

(D) The sequence is arithmetic and can be represented by the function

f(n + 1) = f(n)3/8.

Answer:

Option B:

The sequence is recursive and can be represented by the function

f(n + 1) = f(n) + \frac{3}{8}  .

Explanation:

A sequence of numbers are

\frac{3}{8}, \frac{3}{4}, 1\frac{1}{8}, 1\frac{1}{2}, 1\frac{7}{8}

Let us first change mixed fraction into improper fraction.

\frac{3}{8}, \frac{3}{4}, \frac{9}{8}, \frac{3}{2}, \frac{15}{8}

To find the pattern of the sequence.

To find the common difference between the sequence of numbers.

\frac{3}{4}-\frac{3}{8}=\frac{6}{8}-\frac{3}{8}= \frac{3}{8}

\frac{9}{8}-\frac{3}{4}=\frac{9}{8}-\frac{6}{8}= \frac{3}{8}

\frac{3}{2}-\frac{9}{8}=\frac{12}{8}-\frac{9}{8}= \frac{3}{8}

\frac{15}{8}-\frac{3}{2}=\frac{15}{8}-\frac{12}{8}= \frac{3}{8}

Therefore, the common difference of the sequence is 3.

That means each term is obtained by adding \frac{3}{8} to the previous term.

Hence, the sequence is recursive and can be represented by the function f(n + 1) = f(n) + \frac{3}{8}  .

3 0
3 years ago
PLEASE HELP I NEED THIS ASAP<br> Find the vertex of y=x^2-7x+4
Goshia [24]

Answer:

\mathrm{Minimum}\space\left(\frac{7}{2},\:-\frac{33}{4}\right)

Step-by-step explanation:

y=x^2-7x+4\\\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}\\\mathrm{The\:parabola\:params\:are:}\\a=1,\:b=-7,\:c=4\\x_v=-\frac{b}{2a}\\x_v=-\frac{\left(-7\right)}{2\cdot \:1}\\\mathrm{Simplify}\\x_v=\frac{7}{2}\\\mathrm{Plug\:in}\:\:x_v=\frac{7}{2}\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}\\y_v=\left(\frac{7}{2}\right)^2-7\cdot \frac{7}{2}+4\\

\mathrm{Simplify\:}\left(\frac{7}{2}\right)^2-7\cdot \frac{7}{2}+4:\quad -\frac{33}{4}\\y_v=-\frac{33}{4}\\Therefore\:the\:parabola\:vertex\:is\\\left(\frac{7}{2},\:-\frac{33}{4}\right)\\\mathrm{If}\:a0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}\\a=1\\\mathrm{Minimum}\space\left(\frac{7}{2},\:-\frac{33}{4}\right)

6 0
3 years ago
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