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3 years ago

4. Compare using <> or = -5 O -9

Mathematics

1 answer:

tatiyna3 years ago

7 0

Answer:

-9 <-5 <0

Step-by-step explanation:

hope it helps

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In the fraction 7/8, the numerator denominator is 7.

ololo11 [35]

7/8 is the fraction

7 is the numerator

8 Is the denominator

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3 years ago

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Consider the following.3x + 3y = 8(a) Find y' by implicit differentiation.(b) Solve the equation explicitly for y and differenti

ludmilkaskok [199]

Answer:

a.y'=-1

b.y'=-1

c.Yes

Step-by-step explanation:

We are given that consider a function

$3x+3y=8$

Implicit function: That function is a relation in which dependent variable can not be expressed in terms of independent variable

Explicit function: It is that function in which dependent variable can be expressed in terms of independent variable.

a. $3x+3y=8$

Differentiate w.r.t x then we get

$3+3\frac{dy}{dx}=0$

$3\frac{dy}{dx}=-3$

$\frac{dy}[dx}=\frac{-3}{3}=-1$

$\frac{dy}{dx}=y'=-1$

b. $3x+3y=8$

$3y=8-3x$

$y=\frac{8-3x}{3}$

Differentiate w.r.t x then we get

$\frac{dy}{dx}=\frac{-3}{3}=-1$

$\frac{dy}{dx}=y'=-1$

When we substituting the value of y obtained from part b into a solution of part a then we get

$y'=-1$

Hence, solutions are consistent.

8 0

3 years ago

Simplify each expression. -4x + 1 + 6 - (-9x)

IceJOKER [234]

Step-by-step explanation:

-4x+1+6-(-9x)

-4x+7-(-9x)

5x+7

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3 years ago

A rectangle has an area of 19.38 cm2. When both the length and width of the rectangle are increased by 1.50 cm, the area of the

Lostsunrise [7]

Answer: $5.7\ cm$

Step-by-step explanation:

Given

Rectangle has an area of $19.38\ cm^2$

Suppose rectangle length and width are $l$ and $w$

If each side is increased by $1.50\ cm$

Area becomes $A_2=35.28\ cm^2$

We can write

$\Rightarrow lw=19.38\quad \ldots(i)\\\\\Rightarrow (l+1.5)(w+1.5)=35.28\\\Rightarrow lw+1.5(l+w)+1.5^2=35.28\\\text{use (i) for}\ lw\\\Rightarrow 19.38+1.5(l+w)=35.28-2.25\\\Rightarrow l+w=9.1\quad \ldots(ii)$

Substitute the value of width from (ii) in equation (i)

$\Rightarrow l(9.1-l)=19.38\\\Rightarrow l^2-9.1l+19.38=0\\\\\Rightarrow l=\dfrac{9.1\pm\sqrt{(-9.1)^2-4(1)(19.38)}}{2\times 1}\\\\\Rightarrow l=\dfrac{9.1\pm\sqrt{5.29}}{2}\\\\\Rightarrow l=\dfrac{9.1\pm2.3}{2}\\\\\Rightarrow l=3.4,\ 5.7$