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ollegr [7]
3 years ago
10

Please help me with these two question

Mathematics
1 answer:
Archy [21]3 years ago
6 0
 i don't really know what it is ..umm
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I used a calculator and have solved it without. I didn't get any of these answers. HELP
Andreas93 [3]

Answer:

-9 + 7√5.

Step-by-step explanation:

(3 - √5)(2 + 3√5)

= 3(2 + 3√5) - √5(2 + 3√5))

= 6 + 9√5 - 2√5 - 15

= -9 + 7√5.

6 0
2 years ago
The area of a square flowerbed is 1.21 square yards. What is the length of each side?
gregori [183]

Step-by-step explanation:

area of a square=a*a=a^2

1.21=a^2

a=√1.21

a=1.1yards

6 0
3 years ago
Use the zero product property to find the solutions to the equation x2 + x – 30 = 12.
Butoxors [25]

Answer:

x= 6, -7

Step-by-step explanation:

3 0
2 years ago
The graph of h(x) is shown. Graph of h of x that begins in quadrant two and decreases rapidly following the vertical line which
fgiga [73]

Answer:

<u>x-intercept</u>

The point at which the curve <u>crosses the x-axis</u>, so when y = 0.

From inspection of the graph, the curve appears to cross the x-axis when x = -4, so the x-intercept is (-4, 0)

<u>y-intercept</u>

The point at which the curve <u>crosses the y-axis</u>, so when x = 0.

From inspection of the graph, the curve appears to cross the y-axis when y = -1, so the y-intercept is (0, -1)

<u>Asymptote</u>

A line which the curve gets <u>infinitely close</u> to, but <u>never touches</u>.

From inspection of the graph, the curve appears to get infinitely close to but never touches the vertical line at x = -5, so the vertical asymptote is x = -5

(Please note:  we cannot be sure that there is a horizontal asymptote at y = -2 without knowing the equation of the graph, or seeing a larger portion of the graph).

5 0
2 years ago
Read 2 more answers
Verify the identity. 4 csc 2x = 2 csc2x tan x
vlada-n [284]

Step-by-step explanation:

4csc(2x) = 2csc^2(x) tan(x)

We start with Left hand side

We know that csc(x) = 1/ sin(x)

So csc(2x) is replaced by 1/sin(2x)

4 \frac{1}{sin(2x)}

Also we use identity

sin(2x) = 2 sin(x) cos(x)

4 \frac{1}{2sin(x)cos(x)}

4 divide by 2 is 2

Now we multiply top and bottom by sin(x) because we need tan(x) in our answer

2\frac{1*sin(x)}{sin(x)cos(x)*sin(x)}

2\frac{sin(x)}{sin^2(x)cos(x)}

2\frac{1}{sin^2(x)} \frac{sin(x)}{cos(x)}

We know that sinx/ cosx = tan(x)

Also  1/ sin(x)= csc(x)

so it becomes 2csc^2(x) tan(x) , Right hand side

Hence verified



6 0
3 years ago
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