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3 years ago

Evalua cada expresion si a=10 b=4 c=8 3b^2+c

Mathematics

2 answers:

ASHA 777 [7]3 years ago

6 0

The answer to that is 56
3(4^2)+8=56

FrozenT [24]3 years ago

4 0

Puede que tenga lo hizo mal, pero creo que es 44

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tekilochka [14]

The answer will be -1.7

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3 years ago

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Select the correct answer. Each of the four angles of a quadrilateral measures 90°. How many rectangles can you construct using

professor190 [17]

Answer:

he defenition of a rectangle is that it has 4 angles that measure 90 degrees

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3 years ago

These are Questions/answers for #1, 2, 3, 17, 18, 19 & 20 if you can't see what it says in the picture (Please answer all qu

kkurt [141]

Answer:

See below for answers and explanations (along with a graph for #3)

Step-by-step explanation:

Problem #1

Applying scalar multiplication, $-4w=-4\langle-96,-180\rangle=\langle384,720\rangle$ .

Its magnitude would be $||-4w||=\sqrt{384^2+720^2}=816$ .

Its direction would be $\displaystyle\theta=tan^{-1}\biggr(\frac{720}{384}\biggr)\approx61.927^\circ\approx62^\circ$ .

Thus, B) 816; 62° is the correct answer

Problem #2

Find the time it takes for the ball to cover 13ft:

$x=(24\cos48^\circ)t\\13=(24\cos48^\circ)t\\t\approx0.8095$

Find the height of the ball at the time it takes for the ball to cover 13ft:

$y=6.1+(24\sin48^\circ)t-16t^2\\y=6.1+(24\sin48^\circ)(0.8095)-16(0.8095)^2\\y\approx10.053$

Thus, A) 10.053 is the correct answer

Problem #3

We have $u=\langle0,-8\rangle$ and $v=\langle6,0\rangle$ as our vectors. Thus, $u+v=\langle0+6,-8+0\rangle=\langle6,-8\rangle$ . Attached below is the correct graph. You can also solve the problem visually by using the parallelogram method where the resultant vector is the diagonal of the parallelogram.

Problem 4 (#7)

$\displaystyle t \cdot v=(7)(-10)+(-3)(-8)=-70+24=-46$

Thus, C) -46 is the correct answer

Problem 5 (#8)

Find $r$ and $\theta$ :

$r=\sqrt{x^2+y^2}=\sqrt{2^2+(-8)^2}=\sqrt{4+64}=\sqrt{68}=2\sqrt{17}\approx8.246$

$\displaystyle\theta=tan^{-1}\biggr(\frac{y}{x}\biggr)=tan^{-1}\biggr(\frac{-8}{2}\biggr)\approx-75.964^\circ$

Find the true direction angle accounting for Quadrant IV:

$\theta=360^\circ-75.964^\circ=284.036^\circ$

Write the complex number in polar/trigonometric form:

$z=8.246(\cos284.036^\circ+i\sin284.036^\circ)$

Thus, C) 8.246(cos 284.036° + i sin 284.036°) is the correct answer

Problem 6 (#12)

Eliminate the parameter and find the rectangular equation:

$x=3t\\\frac{x}{3}=t\\ \\y=t^2+5\\y=(\frac{x}{3})^2+5\\y=\frac{x^2}{9}+5\\9y=x^2+45\\0=x^2-9y+45\\x^2-9y+45=0$

Thus, D) x^2-9y+45=0 is the correct answer

Problem 7 (#13)

Find the magnitude of the vector:

$||v||=\sqrt{(-77)^2+36^2}=85$

Find the true direction of the vector accounting for Quadrant II:

$\displaystyle \theta=tan^{-1}\biggr(\frac{36}{-77}\biggr)\approx-25^\circ=180^\circ-25^\circ=155^\circ$

Write the vector in trigonometric form:

$w=85\cos155^\circ i+85\sin155^\circ j$

Thus, D) w=85cos155°i+85sin155°j is the corrwect answer

Problem 8 (#15)

$\frac{z_1+z_2}{2}=\frac{(3-7i)+(-9-19i)}{2}=\frac{-6-26i}{2}=-3-13i=(-3,-13)$

Thus, C) (-3,-13) is the correct answer

Problem 9 (#16)

Treat the golf ball and wind as vectors:

$u=\langle1.3\cos140^\circ,1.3\sin140^\circ\rangle$ <-- Golf Ball

$v=\langle1.2\cos50^\circ,1.2\sin50^\circ\rangle$ <-- Wind

Add the vectors:

$u+v=\langle1.3\cos140^\circ+1.2\cos50^\circ,1.3\sin140^\circ+1.2\sin50^\circ\rangle\approx\langle-0.225,1.755\rangle$

Find the magnitude of the resultant vector:

$||u+v||=\sqrt{(-0.225)^2+1.755^2}\approx1.769$

Find the true direction of the resultant vector accounting for Quadrant II:

$\displaystyle \theta=\tan^{-1}\biggr(\frac{1.755}{-0.225}\biggr)\approx-82.694^\circ\approx-83^\circ=180^\circ-83^\circ=97^\circ$

Thus, B) 1.769 m/s; 97° is the correct answer

Problem 10 (#17)

Identify the vectors and add them:

$u+v+w=\langle50\cos20^\circ,50\sin20^\circ\rangle+\langle13\cos90^\circ,13\sin90^\circ\rangle+\langle35\cos280^\circ,35\sin280^\circ\rangle=\langle50\cos20^\circ+13\cos90^\circ+35\cos280^\circ,50\sin20^\circ+13\sin90^\circ+35\sin280^\circ\rangle=\langle53.062,-4.367\rangle$

Find the magnitude of the resultant vector:

$||u+v+w||=\sqrt{53.062^2+(-4.367)^2}\approx53.241$

Find the true direction of the resultant vector accounting for Quadrant IV:

$\displaystyle \theta=\tan^{-1}\biggr(\frac{-4.367}{53.062}\biggr)\approx-4.705^\circ\approx-5^\circ=360^\circ-5^\circ=355^\circ$

Thus, A) 53.241, 355° is the correct answer

Problem 11 (#18)

We observe that $z_1=-8-6i$ and $z_2=4-4i$ , hence, $z_1+z_2=(-8-6i)+(4-4i)=-4-10i$

Thus, Q is the correct answer

Problem 12 (#19)

Find the dot product of the vectors:

$F_1\cdot F_2=(12000*14500)+(7000*-5000)=174000000+(-35000000)=139000000$

Find the magnitude of each vector:

$||F_1||=\sqrt{12000^2+7000^2}=1000\sqrt{193}\\||F_2||=\sqrt{14500^2+(-5000)^2}=500\sqrt{941}$

Find the angle between the two vectors:

$\displaystyle \theta=\cos^{-1}\biggr(\frac{F_1\cdot F_2}{||F_1||||F_2||}\biggr)\\ \theta=\cos^{-1}\biggr(\frac{139000000}{(1000\sqrt{193})(500\sqrt{941})}\biggr)\\\theta\approx49.282^\circ\approx49^\circ$

Thus, C) 49° is the correct answer

Problem 13 (#20)

Using scalar multiplication, $7v-2w=7\langle8,-3\rangle-2\langle-12,4\rangle=\langle56,-21\rangle-\langle-24,8\rangle=\langle56-(-24),-21-8\rangle=\langle80,-29\rangle=80i-29j$

Thus, A) -80i - 29j is the correct answer

6 0

2 years ago

12 POINT HURRY HURYRJXNXNDJDJDNDN

padilas [110]

Answer:

The fourth choice, Pink, is the correct answer.

Step-by-step explanation:

The first two options intersect, therefore the intersection is the solution. The third set of lines may seem like one line, however since we are dealing with two linear equations, it can be presumed that those are two lines on top of one another, therefore it has infinite solutions. The last option remaining is Pink, in which the lines are parallel and therefore will never intersect, making them have no solution.

8 0

3 years ago

Solve: 1/3a^2-1/a=1/6a^2

Lady bird [3.3K]

Step-by-step explanation:

there are two answers for a

6 0

3 years ago