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2 years ago

Solve the inequality. 4m – 3(m + 1) – 4 < 0 A) m < 7 B) m > 7 C) m < -1 D) m < -7

Mathematics

2 answers:

goldenfox [79]2 years ago

5 0

Answer:

A. M<7

Step-by-step explanation:

Solve for m, by simplifying both sides of the inequality, then isolating the variable.

Inequality Form:

m < 7

Interval Notation:

( − ∞ , 7 )

agasfer [191]2 years ago

3 0

Answer:

A) m < 7

Step-by-step explanation:

4m – 3(m + 1) – 4 < 0

Distribute the 3

4m -3m -3-4<0

m-7 <0

Add 7 to each side

m-7+7 <0+7

m<7

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Given f (x) = 5x + 1, find f(0).

Andrej [43]

Answer:

Step-by-step explanation:

5(0)+1=1 because 5x0=0 then add the 1

7 0

2 years ago

What is .52,57% and 2/5 from least to greatest

Alekssandra [29.7K]

Concert 2/5 into a percent 5*20 = 100 (denominator) so 2* 20 =40
40/100 is 40%
I'm guessing you accidentally put the dor in front of 52 so it would be 40 at least and 52.57 at greatest

5 0

3 years ago

Perform the indicated opera<br> 6) g(n)= n2 - 1<br> h(n)= 4n+1<br> Find (g. h)n)

DaniilM [7]

$(g.h)(n)\,\,is\,\,4n^3+n^2-4n-1\\$

Step-by-step explanation:

We are given:

$g(n)=n^2-1\\h(n)=4n+1$

We need to find (g.h)(n)

Finding (g.h)(n)

$(g.h)(n)=g(n).h(n)\\ (g.h)(n)=(n^2-1)(4n+1)\\ (g.h)(n)=n^2(4n+1)-1(4n+1)\\ (g.h)(n)=4n^3+n^2-4n-1\\$

So, $(g.h)(n)\,\,is\,\,4n^3+n^2-4n-1\\$

Keywords: Composite Function

Learn more about Composite Function at:

#learnwithBrainly

6 0

2 years ago

Evaluate the double integral.

Fynjy0 [20]

Answer:

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

Step-by-step explanation:

The equation of the line through the point $(x_o,y_o)$ & $(x_1,y_1)$ can be represented by:

$y-y_o = m(x - x_o)$

Making m the subject;

$m = \dfrac{y_1 - y_0}{x_1-x_0}$

∴

we need to carry out the equation of the line through (0,1) and (1,2)

i.e

y - 1 = m(x - 0)

y - 1 = mx

where;

$m= \dfrac{2-1}{1-0}$

m = 1

Thus;

y - 1 = (1)x

y - 1 = x ---- (1)

The equation of the line through (1,2) & (4,1) is:

y -2 = m (x - 1)

where;

$m = \dfrac{1-2}{4-1}$

$m = \dfrac{-1}{3}$

∴

$y-2 = -\dfrac{1}{3}(x-1)$

-3(y-2) = x - 1

-3y + 6 = x - 1

x = -3y + 7

Thus: for equation of two lines

x = y - 1

x = -3y + 7

i.e.

y - 1 = -3y + 7

y + 3y = 1 + 7

4y = 8

y = 2

Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7

∴

$\iint_D 8y^2 \ dA = \int^2_1 \int ^{-3y+7}_{y-1} \ 8y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \int ^{-3y+7}_{y-1} \ y^2 \ dxdy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( \int^{-3y+7}_{y-1} \ dx \bigg) dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [xy^2]^{-3y+7}_{y-1} \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( [y^2(-3y+7-y+1)]\bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ([y^2(-4y+8)] \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \int^2_1 \bigg ( -4y^3+8y^2 \bigg ) \ dy$

$\iint_D 8y^2 \ dA =8 \bigg [\dfrac{ -4y^4}{4}+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -y^4+\dfrac{8y^3}{3} \bigg ]^2_1$

$\iint_D 8y^2 \ dA =8 \bigg [ -2^4+\dfrac{8(2)^3}{3} + 1^4- \dfrac{8\times (1)^3}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -16+\dfrac{64}{3} + 1- \dfrac{8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{64-8}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ -15+ \dfrac{56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{-45+56}{3}\bigg]$

$\iint_D 8y^2 \ dA =8 \bigg [ \dfrac{11}{3}\bigg]$

$\iint_D 8y^2 \ dA = \dfrac{88}{3}$

4 0

2 years ago

Write perimeter and area of circle

alexira [117]

Answer:

The perimeter of a circle is π × d. Here's is the diameter of the circle. Hence, the perimeter of a circle is half of that of the circle that is ½ π × d.

The area of a circle is pi times the radius squared (A = π r²). Learn how to use this formula to find the area of a circle when given the diameter.

Step-by-step explanation:

8 0

3 years ago