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djyliett [7]
3 years ago
6

0.00777x4.5/270,000 write in scientific notation

Mathematics
1 answer:
iris [78.8K]3 years ago
4 0

Answer:

87 {6 \frac{?}{?} ?}^{2}

multi

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Triangle J is shown below. James drew a scaled version of Triangle J using a scale factor of 4 and labeled it
Novay_Z [31]

Answer:

The area of triangle K is 16 times greater than the area of triangle J

Step-by-step explanation:

we know that

If Triangle K is a scaled version of Triangle J

then

Triangle K and Triangle J are similar

If two triangles are similar, then the ratio of its areas is equal to the scale factor squared

Let

z -----> the scale factor

Ak ------> the area of triangle K

Aj -----> the area of triangle J

so

z^{2}=\frac{Ak}{Aj}

we have

z=4

substitute

4^{2}=\frac{Ak}{Aj}

16=\frac{Ak}{Aj}

Ak=16Aj

therefore

The area of triangle K is 16 times greater than the area of triangle J

4 0
2 years ago
Read 2 more answers
Help please image is uploaded view please.
harina [27]

Answer:

23

Step-by-step explanation:

we are given that angle NRQ is 78 degrees

we can see from the figure that the sum of the given angles is angle NRQ

so

(8x + 7) + (4x - 1) = 78

12x + 6 = 78

12x = 72

x = 6

Now, we have to find angle PRQ

replacing x with 6 in the equation of angle PRQ

PRQ = 4(6) - 1

PRQ = 23

6 0
3 years ago
Can someone answer this truthfully
juin [17]

Answer:

1- 4x -7 =29, x= 9

2-x/3 - 8 = 12, x=60

Step-by-step explanation:

4x-7 =29

4x= 29 +7

4x= 36

x= 9

x/3 -8 =12

x/3 = 12+ 8

x/3 = 20

x= 20(3)

x= 60

4 0
2 years ago
4x + 11y = -3 and -6x = 18y - 6 (substitution method)
dolphi86 [110]

Answer:

-3 + -6= -9

because there is answer given

6 0
3 years ago
|. Identify the following Pōints of each values.Write your ans
Dmitry_Shevchenko [17]
<h2>✒️VALUE</h2>

\\ \quad  \begin{array}{c} \qquad \bold{Distance \: \green{ Formula:}}\qquad\\ \\ \boldsymbol{ \tt d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}} \end{array}\\  \begin{array}{l} \\ 1.)\: \bold{Given:}\: \begin{cases}\tt D(- 5,6), E(2.-1),\textsf{ and }F(x,0) \\ \tt DF = EF \end{cases} \\ \\  \qquad\bold{Required:}\:\textsf{ value of }x \\ \\ \qquad \textsf{Solving for }x, \\ \\  \tt  \qquad DF = EF \\ \\  \implies\small \tt{\sqrt{(x -(- 5))^2 + (0 - 6)^2} = \sqrt{(x - 2)^2 + (0 - (-1))^2}} \\ \\   \implies\tt\sqrt{(x + 5)^2 + 36 } = \sqrt{(x - 2)^2 + 1 } \\ \\ \textsf{Squaring both sides yields} \\ \\  \implies\tt (x + 5)^2 + 36 = (x - 2)^2 + 1 \\ \\  \implies\tt x^2 + 10x + 25 + 36 = x^2 - 4x + 4 + 1 \\ \\ \implies \tt x^2 + 10x + 61 = x^2 - 4x + 5 \\ \\   \implies\tt10x + 4x = 5 - 61 \\ \\   \implies\tt14x = -56 \\ \\  \implies \red{\boxed{\tt x = -4}}\end{array}  \\  \\  \\  \\\begin{array}{l} \\ 2.)\: \bold{Given:}\: \begin{cases}\tt P(6,-1), Q(-4,-3),\textsf{ and }R(0,y) \\ \tt PR = QR \end{cases} \\ \\ \bold{Required:}\:\textsf{ value of }y \\ \\  \qquad\textsf{Solving for }y, \\ \\  \qquad\tt PR = QR \\ \\  \implies \tt\small{\sqrt{(0 - 6)^2 + (y - (-1))^2} = \sqrt{(0 - (-4))^2 + (y - (-3))^2}} \\ \\   \implies\tt\sqrt{36 + (y + 1)^2} = \sqrt{16 + (y + 3)^2 } \\ \\ \textsf{Squaring both sides yields} \\ \\  \implies \tt \: 36 + (y + 1)^2 = 16 + (y + 3)^2 \\ \\  \implies\tt 36 + y^2 + 2y + 1 = 16 + y^2 + 6y + 9 \\ \\  \implies \tt \: y^2 + 2y + 37 = y^2 + 6y + 25 \\ \\  \implies \tt \: 2y - 6y = 25 - 37 \\ \\ \implies \tt -4y = -12 \\ \\   \implies\red{\boxed{ \tt y = 3}} \end{array}  \\  \\  \\ \begin{array}{l} \\ 3.)\: \bold{Given:}\: \begin{cases}\: A(4,5), B(-3,2),\textsf{ and }C(x,0) \\ \: AC = BC \end{cases} \\ \\ \bold{Required:}\:\textsf{ value of }x \\ \\  \qquad\textsf{Solving for }x, \\ \\   \qquad\tt AC = BC \\ \\ \implies\tt\small{\sqrt{(x - 4)^2 + (0 - 5)^2} = \sqrt{(x - (-3))^2 + (0 - 2)^2}} \\ \\ \implies\tt\sqrt{(x - 4)^2 + 25} = \sqrt{(x + 3)^2 + 4} \\ \\ \textsf{Squaring both sides yields} \\ \\ \implies\tt\:(x - 4)^2 + 25 = (x + 3)^2 + 4 \\ \\ \implies\tt\:x^2 - 8x + 16 + 25 = x^2 + 6x + 9 + 4 \\ \\ \implies\tt\:x^2 - 8x + 41 = x^2 + 6x + 13 \\ \\ \implies\tt-8x - 6x = 13 - 41 \\ \\\implies\tt -14x = -28 \\ \\ \implies\red{\boxed{\tt\:x = 2}} \end{array}

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#BrainlyMathKnower

#5-MinutesAnswer

7 0
2 years ago
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