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3 years ago

48d + 96g = ( d + 2g) need answer plz.........................................

Mathematics

2 answers:

kolbaska11 [484]3 years ago

6 0

I am pretty sure D= -2g

^o^

sashaice [31]3 years ago

5 0

48d + 96g = d + 2g

d+ 2g = 48(d+2g) = 48g + 96g

48d + 96g = 48d + 96g
48. 48
d + 96g = d + 96g
96. 96
d + g = d+ g

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A pizza parlor charges $7.90 for a large pizza plus $1.25 for each additional topping. Write and solve an equation to find the n

zheka24 [161]

Answer:

3 toppings

Step-by-step explanation:

11.65

-7.90

3.75

3.75/1.25=3 toppings

4 0

3 years ago

Three fourths of a number p less subtracted by 2 is 5 sixths of a number r plus 5

goldenfox [79]

Three fourths of a number p less subtracted by 2 is 5 sixths of a number r plus 5

3/4 of p subtracted by 2 is 5/6 of a r plus 5

3/4 of p - 2 is 5/6 of r + 5

$\frac{3}{4} p-2= \frac{5}{6} r+5$ <--- answer?

$\frac{3}{4} p= \frac{5}{6} r+7$

$\frac{3p}{4} = \frac{5r}{6} +7$

$\frac{3p}{4}* \frac{6}{5r} = 7$

$\frac{18p}{20r} = 7$

$\frac{9p}{10r} = 7$

$\frac{p}{r} = 7* \frac{10}{9}$

$\frac{p}{r} = \frac{70}{9}$ <-- answer?

Hope this helps

5 0

3 years ago

M is the midpoint of AB. Find the coordinates of B given A(3,8) and M(5,4).

Harlamova29_29 [7]

Answer: B(7,0)

Step-by-step explanation:

Since coordinate of A(3,8) and M(5,4), where M is the midpoint of AB.

Let the coordinate of B be (x,y)

Therefore, the distance AM is equal to distance MB.

Distance = sqrt{ (5 - 3)² + (4-8)² } =sqrt { (x-5)² + (y-4)²}

Comparing corresponding coordinates,

5-3 = x-5

x= 5+5-3 = 7

4-8 = y-4

y= 4-8+4= 0

Therefore, the coordinate of B is (7,0)

3 0

3 years ago

A wall is 15 ft. High and 10 ft. From a house. Find the length of the shortest ladder which will just touch the top of the wall

Talja [164]

Answer:

the length of the shortest ladder is 11.41 feet

Step-by-step explanation:

The computation of the length of the shortest ladder is shown below:

AB^2 = AC^2 + BC^2

AB^2= (10)^2 + (5.5)^2

AB^2= 100 + 30.25

AB^2 = 130.25

AB = 11.41 feet

hence, the length of the shortest ladder is 11.41 feet

7 0

3 years ago

1) Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given

neonofarm [45]

Answer:

Check below, please

Step-by-step explanation:

Hello!

1) In the Newton Method, we'll stop our approximations till the value gets repeated. Like this

$x_{1}=2\\x_{2}=2-\frac{f(2)}{f'(2)}=2.5\\x_{3}=2.5-\frac{f(2.5)}{f'(2.5)}\approx 2.4166\\x_{4}=2.4166-\frac{f(2.4166)}{f'(2.4166)}\approx 2.41421\\x_{5}=2.41421-\frac{f(2.41421)}{f'(2.41421)}\approx \mathbf{2.41421}$

2) Looking at the graph, let's pick -1.2 and 3.2 as our approximations since it is a quadratic function. Passing through theses points -1.2 and 3.2 there are tangent lines that can be traced, which are the starting point to get to the roots.

We can rewrite it as: $x^2-2x-4=0$

$x_{1}=-1.1\\x_{2}=-1.1-\frac{f(-1.1)}{f'(-1.1)}=-1.24047\\x_{3}=-1.24047-\frac{f(1.24047)}{f'(1.24047)}\approx -1.23607\\x_{4}=-1.23607-\frac{f(-1.23607)}{f'(-1.23607)}\approx -1.23606\\x_{5}=-1.23606-\frac{f(-1.23606)}{f'(-1.23606)}\approx \mathbf{-1.23606}$

As for

$x_{1}=3.2\\x_{2}=3.2-\frac{f(3.2)}{f'(3.2)}=3.23636\\x_{3}=3.23636-\frac{f(3.23636)}{f'(3.23636)}\approx 3.23606\\x_{4}=3.23606-\frac{f(3.23606)}{f'(3.23606)}\approx \mathbf{3.23606}\\$

3) Rewriting and calculating its derivative. Remember to do it, in radians.

$5\cos(x)-x-1=0 \:and f'(x)=-5\sin(x)-1$

$x_{1}=1\\x_{2}=1-\frac{f(1)}{f'(1)}=1.13471\\x_{3}=1.13471-\frac{f(1.13471)}{f'(1.13471)}\approx 1.13060\\x_{4}=1.13060-\frac{f(1.13060)}{f'(1.13060)}\approx 1.13059\\x_{5}= 1.13059-\frac{f( 1.13059)}{f'( 1.13059)}\approx \mathbf{ 1.13059}$

For the second root, let's try -1.5

$x_{1}=-1.5\\x_{2}=-1.5-\frac{f(-1.5)}{f'(-1.5)}=-1.71409\\x_{3}=-1.71409-\frac{f(-1.71409)}{f'(-1.71409)}\approx -1.71410\\x_{4}=-1.71410-\frac{f(-1.71410)}{f'(-1.71410)}\approx \mathbf{-1.71410}\\$

For x=-3.9, last root.

$x_{1}=-3.9\\x_{2}=-3.9-\frac{f(-3.9)}{f'(-3.9)}=-4.06438\\x_{3}=-4.06438-\frac{f(-4.06438)}{f'(-4.06438)}\approx -4.05507\\x_{4}=-4.05507-\frac{f(-4.05507)}{f'(-4.05507)}\approx \mathbf{-4.05507}\\$

5) In this case, let's make a little adjustment on the Newton formula to find critical numbers. Remember their relation with 1st and 2nd derivatives.

$x_{n+1}=x_{n}-\frac{f'(n)}{f''(n)}$

$f(x)=x^6-x^4+3x^3-2x$

$\mathbf{f'(x)=6x^5-4x^3+9x^2-2}$

$\mathbf{f''(x)=30x^4-12x^2+18x}$

For -1.2

$x_{1}=-1.2\\x_{2}=-1.2-\frac{f'(-1.2)}{f''(-1.2)}=-1.32611\\x_{3}=-1.32611-\frac{f'(-1.32611)}{f''(-1.32611)}\approx -1.29575\\x_{4}=-1.29575-\frac{f'(-1.29575)}{f''(-4.05507)}\approx -1.29325\\x_{5}= -1.29325-\frac{f'( -1.29325)}{f''( -1.29325)}\approx -1.29322\\x_{6}= -1.29322-\frac{f'( -1.29322)}{f''( -1.29322)}\approx \mathbf{-1.29322}\\$

For x=0.4

$x_{1}=0.4\\x_{2}=0.4\frac{f'(0.4)}{f''(0.4)}=0.52476\\x_{3}=0.52476-\frac{f'(0.52476)}{f''(0.52476)}\approx 0.50823\\x_{4}=0.50823-\frac{f'(0.50823)}{f''(0.50823)}\approx 0.50785\\x_{5}= 0.50785-\frac{f'(0.50785)}{f''(0.50785)}\approx \mathbf{0.50785}\\$

and for x=-0.4

$x_{1}=-0.4\\x_{2}=-0.4\frac{f'(-0.4)}{f''(-0.4)}=-0.44375\\x_{3}=-0.44375-\frac{f'(-0.44375)}{f''(-0.44375)}\approx -0.44173\\x_{4}=-0.44173-\frac{f'(-0.44173)}{f''(-0.44173)}\approx \mathbf{-0.44173}\\$

These roots (in bold) are the critical numbers

3 0

3 years ago