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3 years ago

Which of these numbers has the greatest absolute value-3 1 -9 5

1 answer:

3 0

I believe it is the 9, with absolute values the number will always be positive so the absolute value of -9 is 9, which is the greatest out of all the options

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please help me it is URGENT pls the first one is solve and the 2nd one is simplify I will give u lots of points thank you

klasskru [66]

I.) (5x+3)/4-(2x-4)/3=5
Clear fractions:
3·((5x+3)/4)=15x+9
4·((2x-4)/3)=8x-16
15x+9-(8x-16)=5
15x+9-8x+16=5
Combine like terms:
7x+25=5
7x=-20
x=-20/7

II.) (3/11)·(5/6)-(9/12)·(4/3)+(5/13)·(6/15)
Remember PEMDAS
So first multiply:
3/11·5/6=15/66
9/12·4/3=3/3·1/1=3/3=1
5/13·6/15=1/13·6/3=6/39=2/13
(15/66)-1+(2/13)
Combine:
15/66-1/1=15/66-66/66=-51/66
-51/66+2/3=-51/66+44/66=-7/66
Answer: -7/66 :)

7 0

3 years ago

Read 2 more answers

Daniel wanted to know his approximate score on the final exam for his mathematics class. His professor hinted that his score was

frozen [14]

Answer: 101.69

Step-by-step explanation:

Formula we use to find the z-score :-

$z=\dfrac{x-\mu}{\sigma}$ (1)

Given : $\mu=90$

$\sigma=7$

Daniel's z score : z= 1.67

Let x be the raw score for Daniel's exam grade, then substitute the values of $\mu,\ \sigma\ and \ x$ in (1), we get

$1.67=\dfrac{x-90}{7}\\\\\Rightarrow\ x-90=7\times1.67\\\\\Rightarrow\ x=11.69+90=101.69$

Hence, the raw score for Daniel's exam grade = 101.69

3 0

3 years ago

Fr e e p o i n t s a n d b r a i n l i e s t

lana66690 [7]

Answer:

Hiii

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

^^^^ please help :)))

ikadub [295]

2x-1.1/4+x/3=2

2x-1/4+x/3=2

2x-1/4+x/3+1/4=2+1/4

2x+x/3=9/4

3(2x+x/3)=3.94

7x=27/4

7x/7=27/4/7

x=27/8

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7 0

3 years ago

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw

vfiekz [6]

First, complete the square in the equation for the second circle to determine its center and radius:

x ² - 10x + y ² = 0

x ² - 10x + 25 + y ² = 25

(x - 5)² + y ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

x = r cos(θ)

y = r sin(θ)

Then

x ² + y ² = 100 → r ² = 100 → r = 10

x ² - 10x + y ² = 0 → r ² - 10 r cos(θ) = 0 → r = 10 cos(θ)

y = 5 → r sin(θ) = 5 → r = 5 csc(θ)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to y = 5 at the point (0, 5).

Split up the region at 3 angles θ₁, θ₂, and θ₃, which denote the angles θ at which the curves intersect. They are

θ₁ = 0 … … … by solving 10 = 10 cos(θ)

θ₂ = π/6 … … by solving 10 = 5 csc(θ)

θ₃ = 5π/6 … the second solution to 10 = 5 csc(θ)

Then the area of the region is given by a sum of integrals:

$\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)$

$=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta$

To compute the integrals, use the following identities:

sin²(θ) = (1 - cos(2θ)) / 2

cos²(θ) = (1 + cos(2θ)) / 2

and recall that

d(cot(θ))/dθ = -csc²(θ)

You should end up with an area of

$=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta$

$=\boxed{25\sqrt3+\dfrac{125\pi}3}$

We can verify this geometrically:

• the area of the larger circle is 100π

• the area of the smaller circle is 25π

• the area of the circular segment, i.e. the part of the larger circle that is bounded below by the line y = 5, has area 100π/3 - 25√3

Hence the area of the region of interest is

100π - 25π - (100π/3 - 25√3) = 125π/3 + 25√3

as expected.

3 0

3 years ago