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2 years ago

Tegan needs to make 68 more snow blocks to finish

Mathematics

1 answer:

Sergeu [11.5K]2 years ago

5 0

Answer:

m = 68 - c

Step-by-step explanation:

It is given that Tegan needs to make 68 more snow blocks to finish building an igloo.

Let m be the number of blocks Tegan still needs to make and c be the number of completed blocks.

If total number of required block is 68 and c blocks are completed then the remaining blocks is the difference between 68 and c.

Number of blocks Tegan still needs to make = 68 - c

We know that c represent the number of blocks Tegan still needs to make.

m = 68 - c

Therefore, the required equation is m = 68 - c.

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If ZFRO = 60 and RFO = 86 Find the measure of 2 FOG<br> R Help pleaseee

tigry1 [53]

Answer:

149°

Step-by-step explanation:

The sum of two interior angles in a triangle is equal to an exterior angle that's supplementary to the third interior angle

<FRO + <RFO = <FOG

60° + 86 = 146

7 0

2 years ago

Solve irrational equation pls

rusak2 [61]

$\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)$

$
\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}$

3 0

2 years ago

Xiao's teacher asked him to rewrite the sum 60+90 the product of the GCF of the two numbers and a sum. Axial wrote 3(20+30). Wha

LuckyWell [14K]

60 + 90

GCF = 30

(30 * 2) + (30 * 3) = 30(2 + 3) <== the correct way

The mistake Axial made was he did not pick the greatest common factor...he picked 3 instead of 30.

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2 years ago

Read 2 more answers

What is the area?<br> 4.6 yd<br> 7.7 yd

mario62 [17]

Answer:

A = 35.42 yd²

Step-by-step explanation:

We need to find the area if the dimensions are given as 4.6 yd and 7.7 yd.

The area of a rectangle shaped figure is given by :

A = lb

Where

l is length and b is the breadth

So, put all the values,

A = (4.6)(7.7)

= 35.42 yd²

So, the area of the figure is equal to 35.42 yd².

5 0

2 years ago

The length of a rectangle is one inch less than three times its width. If the perimeter of the rectangle is 54 inches, find the

CaHeK987 [17]

w - width

3w - 1 - length

54 in - perimeter

w + w + (3w - 1) + (3w - 1) = 8w - 2 - perimeter

The equation:

8w - 2 = 54 <em>add 2 to both sides</em>

8w = 56 <em>divide both sides by 8</em>

w = 7

3w - 1 = 3(7) - 1 = 21 - 1 = 20

<h3>Answer: The length = 20 in.</h3>

7 0

3 years ago