All categories

1 year ago

if 5 times a number is increased by 4 the result is at least 19. find the least possible number that satisfies these conditions

Mathematics

1 answer:

ludmilkaskok [199]1 year ago

4 0

Answer:

Step-by-step explanation:

5 times 3 is 15

15 plus 4 equals 19

Hope this helps :)

You might be interested in

What is the missing angle

igomit [66]

The missing angle is 53 degrees (82+45=127) (180-127=53)

7 0

2 years ago

Read 2 more answers

Factor the trinomial X2-12x +16

wlad13 [49]

2*&/45/+=67364/3/4/32:5,6544 is hothead answer

6 0

2 years ago

Which of the following expressions are equivalent to -0.5(1.7+1.7)−0.5(1.7+1.7)minus, 0, point, 5, left parenthesis, 1, point, 7

Allisa [31]

The expressions equivalent to -0.5(1.7+1.7) are -0.5(1.7) - 0.5(1.7)

and 2(-0.5(1.7))

<h3>Distributive law of expansion</h3>

Given the expression below;

-0.5 (1.7 + 1.7)

According to the distributive law;

A(B+C) = AB + AC

Expand

-0.5(1.7) + (-0.5)(1.7)

-0.5(1.7) - 0.5(1.7)

2(-0.5(1.7))

Hence the expressions equivalent to -0.5(1.7+1.7) are -0.5(1.7) - 0.5(1.7)

and 2(-0.5(1.7))

Learn more on distributive law here: brainly.com/question/25224410

#SPJ1

7 0

2 years ago

Please see attachment

Dafna11 [192]

Answer:

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

Step-by-step explanation:

Step(i):-

Given function

$f(x) = \frac{-x}{2x^{2} +1}$ ...(i)

Differentiating equation (i) with respective to 'x'

$f^{l} = \frac{2x^{2} +1(-1) - (-x) (4x)}{(2x^{2}+1)^{2} }$ ...(ii)

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} }$

Equating Zero

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

$\frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

$2 x^{2}-1 = 0$

$2 x^{2} = 1$

$x^{2} = \frac{1}{2}$

$x = \frac{-1}{\sqrt{2} } , x = \frac{1}{\sqrt{2} }$

Step(ii):-

Again Differentiating equation (ii) with respective to 'x'

$f^{ll}(x) = \frac{(2x^{2} +1)^{2} (4x) - 2(2x^{2} +1) (4x)(2x^{2}-1) }{(2x^{2}+1)^{4} }$

put

$x = \frac{1}{\sqrt{2} }$

$f^{ll} (x) > 0$

The absolute minimum value at $x = \frac{1}{\sqrt{2} }$

Step(iii):-

The value of absolute minimum value

$f(x) = \frac{-x}{2x^{2} +1}$

$f(\frac{1}{\sqrt{2} } ) = \frac{-\frac{1}{\sqrt{2} } }{2(\frac{1}{\sqrt{2} } )^{2} +1}$

on calculation we get

The value of absolute minimum value = - 0.3536

Final answer:-

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

3 0

2 years ago

Solve for k 8k + 2m = 3m + k

butalik [34]

Answer:

k= m/7

Step-by-step explanation:

7 0

2 years ago