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2 years ago

What number is bigger 10.65 or 1.056

Mathematics

2 answers:

Vladimir [108]2 years ago

5 0

10.65 is bigger because 65 is larger than 56

Mandarinka [93]2 years ago

5 0

10.65 is larger than 1.056

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Lines k and n are perpendicular. If the slope of line k is -6, what is the slope of line n?

CaHeK987 [17]

Lines k and n are perpendicular

The slope of line k is -6

Fact: The product of slopes of two perpendicular lines = -1

So the slope of line n = -1 ÷ -6 = 1/6

The correct choice is 'D.'

8 0

2 years ago

Triangles ABC and EDC are outlined on a bridge. The triangles share vertex C and angles D and B are right angles. A new bridge s

dexar [7]

CHECK THE ATTACHED FIQURE FOR THE BRIDGE

Answer:

60 ft

Step-by-step explanation:

From the question we know that triangles ABC and EDC are in a 1:1 with this given ratio it implies that

triangles ABC and EDC are congruent then we can say

side EC = side AC

3x + 9 = 5x - 5

Then we can simplify to know value of x

3x + 14= 5x

2x = 14

x = 7

But we know that AC= 5x - 5 , then substitute value of x into it

AC = 5x + 5 = 5(7) - 5

= 35 - 5

AC= 30 ft

Also EC= 3x + 9 then substitute value of x into it

EC = 3x + 9 = 3(7) + 9

= 21 + 9

EC= 30 ft

Then the the distance between the top and bottom of the bridge, in feet, = EC+AC

= 30 + 30 = 60 ft

4 0

2 years ago

Here is a diagram and its corresponding equation. Find the solution to the equation and explain your reasoning. (Please help ASA

jeka94

Use order of operations for this

4x+17=23

Move the 17 to the 23

4x=23-17

Then move the 4 over leaving the x by itself

X=6/4

X=1.5

3 0

2 years ago

Read 2 more answers

Help math question derivative!

atroni [7]

Let $f(x)=\sec^{-1}x$ . Then $\sec f(x)=x$ , and differentiating both sides with respect to $x$ gives

$(\sec f(x))'=\sec f(x)\tan f(x)\,f'(x)=1$
$f'(x)=\dfrac1{\sec f(x)\tan f(x)}$

Now, when $x=\sqrt2$ , you get

$(\sec^{-1})'(\sqrt2)=f'(\sqrt2)=\dfrac1{\sec\left(\sec^{-1}\sqrt2\right)\tan\left(\sec^{-1}\sqrt2\right)}$

You have $\sec^{-1}\sqrt2=\dfrac\pi4$ , so $\sec\left(\sec^{-1}\sqrt2\right)=\sqrt2$ and $\tan\left(\sec^{-1}\sqrt2\right)=1$ . So $(\sec^{-1})'(\sqrt2)=\dfrac1{\sqrt2\times1}=\dfrac1{\sqrt2}$

5 0

2 years ago

Where is expression that equals 40 it through the exponent

Lisa [10]

10 it the power of 4

7 0

2 years ago