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2 years ago

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<_c<_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