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3 years ago

Round 0.578 to the nearest tenth

Mathematics

2 answers:

Deffense [45]3 years ago

5 0

0.578 rounded to the nearest tenth would be 0.6, since the 7 in the hundredths spot is higher than 5.

noname [10]3 years ago

3 0

The answer is 0.58 because 8 is a bigger number than five and you always round up when the number is larger than five. Therefore you get 0.58

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3 years ago

Which statement is true?<br> GRAPHA<br> GRAPH B

WARRIOR [948]

Answer:

Graph B is correct

Step-by-step explanation:

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3 years ago

Read 2 more answers

XYZ has coordinates X(2, 3), Y(1,4), and Z(8,9). A translation maps X to X'(4,7). What are the coordinates for Y' and Z' for thi

Rashid [163]

Answer:

The coordinates are $Y'(x,y) = (3, 8)$ and $Z'(x,y) = (10, 13)$ .

Step-by-step explanation:

First, we have to derive an expression for translation under the assumption that each point of XYZ experiments the same translation. Vectorially speaking, translation from X to X' is defined by:

$X'(x,y) = X(x,y) + T(x,y)$ (1)

Where $T(x,y)$ is the vector translation.

If we know that $X(x,y) = (2,3)$ and $X'(x,y) = (4,7)$ , then the vector translation is:

$T(x,y) = X'(x,y)-X(x,y)$

$T(x,y) = (4,7) - (2,3)$

$T(x,y) = (2, 4)$

Then, we determine the coordinates for Y' and Z':

$Y'(x,y) = Y(x,y) + T(x,y)$

$Y'(x,y) = (1,4) + (2,4)$

$Y'(x,y) = (3, 8)$

$Z'(x,y) = Z(x,y) + T(x,y)$

$Z'(x,y) =(8,9) + (2,4)$

$Z'(x,y) = (10, 13)$

The coordinates are $Y'(x,y) = (3, 8)$ and $Z'(x,y) = (10, 13)$ .

7 0

3 years ago

Perform the indicated operation. Be sure the answer is reduced.

avanturin [10]

<h3>Given Equation:-</h3>

$\boxed{ \rm \frac{4x^{2}y^{3}z}{9} \times \frac{45y}{8 {x}^{5} {z}^{5} }}$

<h3>Step by step expansion:</h3>

$\dashrightarrow \sf\dfrac{4x^{2}y^{3}z}{9} \times \dfrac{45y}{8 {x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{ \cancel4x^{2}y^{3}z}{9} \times \dfrac{45y}{ \cancel8 {x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{9} \times \dfrac{45y}{2{x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{ \cancel9} \times \dfrac{ \cancel{45}y}{2{x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{x^{2}y^{3}z}{1} \times \dfrac{5y}{2{x}^{5} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{x^{0}y^{3}z}{1} \times \dfrac{5y}{2{x}^{5 - 2} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{y^{3}z}{1} \times \dfrac{5y}{2{x}^{3} {z}^{3} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{y^{3}z {}^{0} }{1} \times \dfrac{5y}{2{x}^{3} {z}^{3 - 1} }$

$\\ \\$

$\dashrightarrow \sf\dfrac{y^{3}}{1} \times \dfrac{5y}{2{x}^{3} {z}^{2} }$

$\\ \\$

$\dashrightarrow \sf \dfrac{5y \times {y}^{3} }{2{x}^{3} {z}^{2} }$

$\\ \\$

$\dashrightarrow \sf \dfrac{5y {}^{0} \times {y}^{3 + 1} }{2{x}^{3} {z}^{2} }$

$\\ \\$

$\dashrightarrow \sf \dfrac{5 \times {y}^{4} }{2{x}^{3} {z}^{2} }$

$\\ \\$

$\dashrightarrow \bf \dfrac{5 {y}^{4} }{2{x}^{3} {z}^{2} }$

$\\ \\$

$\therefore \underline{ \textbf{ \textsf{option \red c \: is \: correct}}}$

8 0

2 years ago

In triangle ABC, m<A = 8x +12, m<B = 5x+14, m<C = 2x +4. What is the m<A ? Need help....please

Elena L [17]

Triangle
m<A + m<B + m<C = 180
8x +12 + 5x+14 + <span>2x +4 = 180
15x + 30 = 180
15x = 180 -30
15x = 150
x = 10

m<A </span><span>= 8x +12
</span>m<A = 8(10) +12
m<A = 80 + 12
m<A = 92

answer
m<A = 92

6 0

3 years ago