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horsena [70]
3 years ago
14

Find the solution to the system of equations.

Mathematics
1 answer:
Anna [14]3 years ago
5 0
No solution it’s infinite
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Combining transformations result in interesting designs.<br><br> True False
aev [14]
True is the correct answer Hope this helped =)
7 0
4 years ago
Scientists found the wreck of the titanic 3797 m below sea level. How many km below sea level is the wreck?
WINSTONCH [101]

We are given that the titanic is found 3,797 m below sea level. This is simply equivalent to 3.797 km or 3.8 km when rounded off to one decimal.

We should take note that the letter k in km means kilo and a kilo is equivalent to 1000. So this means that for 1 km there is exactly 1000 m. So to convert, simply multiply a value with unit m with this conversion factor. Conversion factor = 1 km / 1000 m

Depth of titanic = 3,797 m (1 km / 1000 m)

Depth of titanic = 3.797 km = 3.8 km

4 0
3 years ago
Read 2 more answers
Find u, v , u , v , and d(u, v) for the given inner product defined on Rn. u = (0, −4), v = (5, 3), u, v = 3u1v1 + u2v2
tigry1 [53]

Answer:

=-12

d(u,v)=2\sqrt{31}

Step-by-step explanation:

We are given that inner product defined on R_n

=3u_1v_1+u_2v_2

u=(0,-4),v=(5,3)

We have to find the value of <u,v> and d(u,v)

We have u_1=0,u_2=-4,v_1=5,v_2=3

Substitute the value then we get

=3(0\cdot5)+(-4)(3)=-12

Now, d(u,v)=\left \|v-u\right \|

Using this formula we get

d(u,v)=\left \| (5,7) \right \|=\sqrt{3(5)^2+(7)^2}=\sqrt{75+49}=\sqrt{124}

d(u,v)=2\sqrt{31}

7 0
3 years ago
Find the measure of side NO. Figures are not drawn to scale.​
miskamm [114]
Its being dilated by 3 therefore your answer is 9.9
7 0
3 years ago
Solve for x in the equation x² + 14x + 17 = - 96
gavmur [86]

Answer:

Solutions of x are;

x = -7 + 8i and x = -7 -8i

Step-by-step explanation:

Given the equation: x^2+14x+17 = -96

Add 96 both sides we get;

x^2+14x+17+96 = 0

x^2+14x+113= 0

Using quadratic formula ax^2+bx+c = 0 then the solution is given by:

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

On comparing we have;

a= 1,  b =14 and c =113

x = \frac{-14\pm\sqrt{(14)^2-4(1)(113)}}{2(1)}

x = \frac{-14\pm\sqrt{196-452}}{2}

or

x = \frac{-14\pm\sqrt{-256}}{2}

Simplify:

x = \frac{-14\pm 16i}{2} ;  where i is the imaginary, i^2= -1

or

x = -7 \pm 8i

Therefore,  the solution of x are;  x = -7 + 8i and x = -7 -8i

4 0
3 years ago
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