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2 years ago

2) Which ordered pairs represent a proportional relationship between x and y?

Mathematics

2 answers:

kirill [66]2 years ago

7 0

Answer

Step-by-step explanation:

2,5 and 6,15

ZanzabumX [31]2 years ago

4 0

Answer:

Step-by-step explanation:

you can multiply the original x and y by 3 and get the second x and y

i hope this helps

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Which pairs of polygons are congruent?

Nadya [2.5K]

I Say C But If It’s Wrong Sorry

7 0

2 years ago

Read 2 more answers

Evaluate 3 - 24 = 8 + 4^2.<br> A 0<br> B 2<br> C 16<br> D 24

oee [108]

Answer:

-21=24

Step-by-step explanation:

Subtract 24 from 3

-21=8+2^2

Raise 4 to the power of 2

-21=8+16

Add 8 and 16

-21=24

7 0

2 years ago

Surface area and volume of a regular triangular pyramid that has a base edge of 16 cm and a slant height of 15 cm

Schach [20]

Answer:

a) Surface area = 616 cm²

b) Volume = V = 1083 cm³

Step-by-step explanation:

Surface area and volume of a regular triangular pyramid that has a base edge of 16 cm and a slant height of 15 cm

a) Formula for surface area = Base area × 1/2(perimeter × slant height)

Base area = Base edge²

=( 16 cm)² = 256 cm²

Perimeter = Base edge × 3

= 16 cm × 3 = 48 cm

Hence:

Surface Area = 256 cm² + 1/2(48 × 15)

= 256 cm² + 360 cm²

= 616 cm²

b) Volume = 1/3 × Base area × Height

1082.758616785cm³

Approximately = 1083 cm³

8 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

I WILL MARK BRAINLIEST

vagabundo [1.1K]

Answer:

$\frac{81}{m^7}$