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Nostrana [21]
3 years ago
13

Solve (images attached) can u check if correct

Mathematics
1 answer:
viva [34]3 years ago
7 0

Answer:

Yes it's correct

Step-by-step explanation:

90 + 40 = 130

180 - 130 = 50

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3x-(2x-1)=7x-(3*5x)+(-x+24)
Paraphin [41]

Answer:

\boxed{\boxed{\sf x=\frac{23}{10}}\: \sf or \:\boxed{x=2.3}}

_________________

\boxed{\sf Step\: By\:Step:- }

\sf 3x-\left(2x-1\right)=7x-\left(3\times \:5x\right)+\left(-x+24\right)

<u>Remove the parentheses:</u>

\to\sf 3x-\left(2x-1\right)=7x-3\times \:5x-x+24

<u>Combine like terms:</u>

\sf ^*7x-x=6x

\to\sf 3x-\left(2x-1\right)=6x-3\times \:5x+24

<u>Multiply 3 and 5x = 15x:-</u>

\to\sf 3x-\left(2x-1\right)=6x-15x+24

<u>Combine like terms:</u>

\sf ^*6x-15x=-9x

\to\sf 3x-\left(2x-1\right)=-9x+24

<u>Expand: 3x-(2x-1)= x+1</u>

\to\sf x+1=-9x+24

<u>Subtract 1 from both sides:</u>

\to\sf x+1-1=-9x+24-1

\to\sf x=-9x+23

<u>Add 9x to both sides:</u>

\to\sf x+9x=-9x+23+9x

\to\sf 10x=23

<u>Divide both sides by 10:</u>

\to\sf \cfrac{10x}{10}=\cfrac{23}{10}

\to\sf x=\cfrac{23}{10}

<u>________________________________</u>

3 0
3 years ago
Evaluate in the following expression when x = 4 and y = 5​
PIT_PIT [208]
Fraction with x squared plus y cubed in the numerator and 2 plus x in the denominator.
3.8
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7 0
3 years ago
Sin(x2 + y2) da r , where r is the region in the first quadrant between the circles with center the origin and radii 2 and 5
sweet [91]
Converting to polar coordinates gives

\displaystyle\iint_R\sin(x^2+y^2)\,\mathrm dA=\int_{\theta=0}^{\theta=\pi/2}\int_{r=2}^{r=5}\sin(r^2)r\,\mathrm dr\,\mathrm d\theta
=\displaystyle\pi\int_{r=2}^{r=5}2r\sin(r^2)\,\mathrm dr
=\displaystyle\pi\int_{r^2=4}^{r^2=25}\sin(r^2)\,\mathrm d(r^2)
=-\cos(r^2)\bigg|_{r^2=4}^{r^2=25}
=\pi(\cos4-\cos25)
3 0
3 years ago
What is the value of the expression?
lora16 [44]
-square root 3 over 3
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7 0
3 years ago
Can someone pls give me the answer I don’t get it and it’s due in 20 mins
horrorfan [7]

Answer:

Step-by-step explanation:

Cameron, Eliot, Ryan and Ava can ride as their height is more than or equal to 54

7 0
3 years ago
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