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tatiyna
3 years ago
9

1.8 × 10^9 ? 8.7 × 10^7 compare A.< b. > c.=

Mathematics
1 answer:
Ede4ka [16]3 years ago
8 0
1.8 x 10^9 = 1,800,000,000
8.7 x 10^7 = 87,000,000

1.8 x 10^9 > 8.7 x 10^7
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which of the following is the equation of the line that passes though the point (-2,1) and has a slope of 4
Ierofanga [76]

An equation that goes through (-2, 1) and has a slope of 4 is y = 4x + 9.


You can find this by looking for the y-intercept (b) by using the slope (m), the point and slope intercept form. The work is below for you.


y = mx + b

1 = 4(-2) + b

1 = -8 + b

9 = b


Now we can use that and the slope to create the equation y = 4x + 9

4 0
3 years ago
Verify identity list steps. Cot(t)(1-cos^2(t))=cos(t)sin(t)
storchak [24]
Remember: We have to work from either the LHS or the RHS.
(Left hand side or the Right hand side)

You should already know this:

\huge{Cot(t) = \frac{1}{tan(t)} = \frac{1}{\frac{sin(t)}{cos(t)}} = 1\div \frac{sin(t)}{cos(t)} = 1\times \frac{cos(t)}{sin(t)}=\boxed{\frac{cos(t)}{sin(t)}}


You should also know this:

sin^2(t) + cos^2(t) = 1\\\\\boxed{sin^2(t)} = 1 - cos^2(t)

So plugging in both of those into our identity, we get:

\frac{cos(t)}{sin(t)}\cdot sin^2(t) = cos(t)\cdot sin(t)

Simplify the denominator on the LHS (Left Hand Side)

We get:

cos(t) \cdot sin(t) = cos(t) \cdot sin(t)

LHS = RHS

Therefore, identity is verified.
4 0
3 years ago
41. Suppose a family drives at an average rate of 60 mi/h on the way to visit relatives and then at an average rate of 40 mi/h o
Afina-wow [57]

Answer:

The answer is below

Step-by-step explanation:

a) Let x represent the time taken to drive to see the relatives and let d be the distance travelled to go, hence:

60 mi/h = d/x

d = 60x

When returning, they still travelled a distance d, since the return trip takes 1 h longer than the trip there, therefore:

40 mi/h = d/(x+1)

d = 40(x + 1) = 40x + 40

Equating both equations:

60x = 40x + 40

60x - 40x = 40

20x = 40

x = 40/20

x = 2 h

The time taken to drive there = x = 2 hours

b) The time taken for return trip = x + 1 = 2 + 1 = 3 hours

c) The distance d = 60x = 60(2) = 120 miles

The total distance to and fro = 2d = 2(120) = 240 miles

The total time to and fro = 2 h + 3 h = 5 h

Average speed = total distance / total time = 240 miles / 5 h = 48 mi/h

5 0
3 years ago
What is the difference between a linear function and a nonlinear function? Explain what each looks like when represented as a ta
Degger [83]

Answer:

<h3>Sample Answer:<em> </em></h3><h3><em>A linear function has a constant rate of change, while a nonlinear function does not. For a table of values to be linear, the outputs must have a constant rate of change as the inputs increase by 1. On a graph, the function must be a straight line to be linear.</em></h3>

<em />

6 0
3 years ago
Read 2 more answers
Help me with this please!!
valina [46]
<h2><u>Given:</u><u>-</u></h2>

  • Points C = (-7,2) → \sf{(X_1,Y_1)}
  • D = (3,12) → \sf{(X_2,Y_2)}

<h2><u>To </u><u>Find</u><u>:</u><u>-</u></h2>

  • The Midpoint of CD.

<h2><u>Required</u><u> </u><u>Response</u><u>:</u><u>-</u></h2>

Let,

Midpoint of CD be (x,y).

WKT,

\boxed{\sf{(x,y) = \frac{X_1+X_2}{2},\frac{Y_1+Y_2}{2}}}

→\;{\sf{\frac{-7+3}{2},\frac{2+12}{2}}}

→\;{\sf{\frac{-4}{2},\frac{14}{2}}}

→\;{\sf{-2,7}}

The Midpoint of CD ◕➜ \Large{\red{\mathfrak{(-2,7)}}}

Let,

The centre be O

Radius = CO & OD

Here, C = (-7,2) → \sf{(X_1,Y_1)}

O = (-2,7) → \sf{(X_2,Y_2)}

\boxed{\sf{Distance = \sqrt{(X_2-X_1)²+(Y_2-Y_1)²}}}

→\;{\sf{\sqrt{(-2+7)²+(7-2)²}}}

→\;{\sf{\sqrt{5²+5²}}}

→\;{\sf{\sqrt{25+25}}}

→\;{\sf{\sqrt{50}}}

→\;{\sf{5√2 (or) 7.07}}

Radius of Circle ◕➜ \Large{\red{\mathfrak{7.07}}}

<h2>Option D.</h2>

Hope It Helps You ✌️

3 0
3 years ago
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