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3 years ago

-353.92- ( -283.56 -131.29 =

Mathematics

2 answers:

sergiy2304 [10]3 years ago

6 0

Answer:

-768.21

Step-by-step explanation:

<em>I hope this helped you! Have a great day!</em>

solniwko [45]3 years ago

5 0

The answer for the problem is -768.21

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Which is the quotient of 5 ÷ 1/4 ? Use the model to help. due TOMMOROW help

DaniilM [7]

The answer would be 1.25.

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3 years ago

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spayn [35]

Answer:

Step-by-step explanation:

Note: Take note , absolute value will give the magnitude of the value. (Which means ignoring the + / - signs.)

Eg. | - 19 | = 19

Therefore,

$\frac{12(30 - (9 + {4}^{2})) }{10 - | - 6| } \\ = \frac{12(30 - (9 + 16))}{10 - 6} \\ = \frac{12(30 - 25)}{4} \\ = \frac{12(5)}{4} \\ = \frac{60}{4} \\ = 15$

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2 years ago

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Step-by-step explanation:

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4 years ago

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Find the angle between the given vectors to the nearest tenth of a degree. u = <-5, -4>, v = <-4, -3> (1 point)

sesenic [268]

Angle between $u = -5i-4j , v=-4i-3j$ is $x =0$ ° .

<u>Step-by-step explanation:</u>

We have , two vectors u = <-5, -4>, v = <-4, -3> or , $u = -5i-4j , v=-4i-3j$

We need to find angle between these two vectors . Let's find out:

We know that dot product of two vectors is defined as :

$u.v =|u|(|v|)cosx$ , where x is angle between u & v !

⇒ $u.v =|u|(|v|)cosx$

⇒ $cosx =\frac{u.v}{|u|(|v|)}$

Now , $u.v = (-5i-4j)(-4i-3j)$

⇒ $u.v = (-5i-4j)(-4i)-(-5i-4j)(3j)$

⇒ $u.v = 20+12$ { $i(j) = j(i) =0$ }

⇒ $u.v = 32$

Now , Modulus of any vector $r = xi+yj$ is $|r| = \sqrt{x^{2}+y^{2}}$ So ,

$|u| = \sqrt{(-5)^{2}+(-4)^{2}} = \sqrt{25+16} = \sqrt{41} \\\\|v| = \sqrt{(-4)^{2}+(-3)^{2}} = \sqrt{16+9} = \sqrt{25} = 5$

Putting all these values in equation $cosx =\frac{u.v}{|u|(|v|)}$ we get:

⇒ $cosx =\frac{32}{5(\sqrt{41})}$

⇒ $cos^{-1}(cosx) =cos^{-1}(\frac{32}{5(6.4)})$

⇒ $x =cos^{-1}(1)$ { $cos0 = 1$ }

⇒ $x =0$ °

Therefore , Angle between $u = -5i-4j , v=-4i-3j$ is $x =0$ ° .

8 0

3 years ago

Use the chain rule calculate dw/dr , dw/ds and dw/dt

Liono4ka [1.6K]

Answer:

$\frac{dw}{dr} = \frac{8\cdot r}{4\cdot r^{2}-t^{2}+5\cdot s^{2}}$ , $\frac{dw}{ds} = \frac{10\cdot s}{4\cdot r^{2}-t^{2}+5\cdot s^{2}}$ , $\frac{dw}{dt} = -\frac{2\cdot t}{4\cdot r^{2}-t^{2}+5\cdot s^{2}}$

Step-by-step explanation:

We proceed to derive each expression by rule of chain. Let be $w = \ln (x+2\cdot y + 3\cdot z)$ , $x = r^{2}+t^{2}$ , $y = s^{2}-t^{2}$ and $z = r^{2}+s^{2}$ :