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mihalych1998 [28]
2 years ago
12

I need help..........

Mathematics
1 answer:
Sedbober [7]2 years ago
6 0
49 would be in the next space after 42 because this sequence is counting by 7s
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2248/98 as a mix fraction?
Anarel [89]
The answer is 22 46/49 
Answer:
= 2248/1 ÷ 98/1

= (2248 × 1) / (98 × 1)

= 2248/98

= 1124/49

<span>= 22 46/49</span>

3 0
3 years ago
Don’t understand this question???
FromTheMoon [43]

Answer:

what that

Step-by-step explanation:

i cant understand

8 0
2 years ago
Help??????????????????????
Bumek [7]
Answer: y= -5/2x+3
Explanation: by converting the equation into slope-intercept form you can use the slope and y-intercept given in the equation to graph.


Step-by-step:
5x= -2y+6
2y= -5x+6
Y= -5/2x+3
Slope= -5/2x
Y-intercept= 3

3 0
3 years ago
Complete the equation of the line through (3,−8) and 6,−4). Use exact numbers. y=
Alisiya [41]

Answer:

hello : y = 4/3x -12

Step-by-step explanation:

Hello : let  A(3,- 8)    B(6,-4)

the slope is :   (YB - YA)/(XB -XA)

(-4+8)/(6-3)  = 4/3

the slope is  :  4/3

y -(-4) = 4/3( x -6)

y + 4 = 4/3x - 8

y = 4/3x -12....(  the equation of the line )

8 0
3 years ago
Read 2 more answers
Show there is a number c ,with 0&lt;_c&lt;_1,such that f(c)=0 for the equation f(x)=x^3+x^2-1
Rainbow [258]

Answer:

Use Mean Value theorem.

Step-by-step explanation:

Statement: If f(x) is continuous on [a, b] and differentiable on (a, b) then there is at least one 'c' (a < c < b), then we have:

                  f'(c) = $ \frac{f(b) - f(a)}{b - a} $

Here, f(x) = x³ + x² - 1. a = 0, b =1

Since, f(x) is a polynomial, it is continuous and differentiable on the interval.

f'(x) = 3x² + 2x

⇒ f'(c) = 3c² + 2c

Using Mean value theorem, we have:

3c² + 2c = $ \frac{f(1) - f(0)}{1 - 0} $

f(1) = 1 + 1 - 1 = 1

f(0) = 0 + 0 - 1 = - 1

$ \implies f'(c) = \frac{1 - (-1)}{1 - 0} $

$ \implies f'(c) = \frac{2}{1} = 2 $

Therefore, we have: 3c² + 2c = 2

Rearranging this, we have: 3c² + 2c - 2 = 0 which is a quadratic equation.

Now, we find the roots of the equation using the formula:

We have: c = $ \frac{- 2 \pm \sqrt{4 - 4(3)(2)}}{2.3} $

= $ \frac{- 2 \pm \sqrt{4 + 24}}{6} $

= $ \frac{- 2 \pm 2\sqrt{7}}{6} $

= $ \frac{- 1 \pm \sqrt{7}}{3} $

The roots are: c = $ \frac{- 1 + \sqrt{7}}{6} , \frac{- 1 - \sqrt{7}}{6} $

Since, our root should lie between 0 and 1, we eliminate $ \frac{- 1 - \sqrt{7}}{6} $.

Hence, the value of c = $ \frac{- 1 + \sqrt{7}}{6} $

So, we have proved the existence of 'c' and have determined the value of 'c' as well.

5 0
2 years ago
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