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Oxana [17]
3 years ago
11

-7 \times (7 + 9)−7×(7+9)

Mathematics
1 answer:
prisoha [69]3 years ago
8 0

Answer:

−224

Step-by-step explanation:

−7(7+9)−7(7+9)

=(−7)(16)−7(7+9)

=−112−7(7+9)

=−112−(7)(16)

=−112−112

=−224

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Which of the following are true statements about any regular polygon
beks73 [17]

Answer:

need

Step-by-step explanation:

4 0
3 years ago
An airplane leaves Los Angeles and travels five hours to New YorkCity. The return trip travels along the same route and takes si
finlep [7]

The distance of the plane flies between the two cities is 1,113.75  miles

The formula for calculating the distance travelled is expressed as:

D = vt where:

v is the speed

t is the time taken

Given the following:

v = 99mi/hr

Total time taken = 5hr + 6.25hr

Total time taken = 11.25hr

Get the required distance

Distance = 99mi/hr * 11.25hr

Distance = 1,113.75miles

Hence the distance of the plane flies between the two cities is 1,113.75  miles

learn more here: brainly.com/question/7245260

4 0
3 years ago
4. Trazá y responde: Juan dibujó un ángulo de 150° y le trazó su bisectriz. Le quedaron dos ángulos, uno
Pachacha [2.7K]

Answer:

no, no esta bien. la bisectriz es la mitad de 150°, entonces seria 75° y 75°

Step-by-step explanation:

4 0
2 years ago
I need to know how to make it in its simplest form
Mrrafil [7]
Multiply denominator by whole number then add that number to the nominator. Which would be 5 times 1 plus 3. SO 5+3 IS 8 and you keep the denominator so the first one is 8/5
6 0
3 years ago
(a) Find the equation of the normal to the hyperbola x = 4t, y = 4/t at the point (8,2).
gtnhenbr [62]

The slope of the tangent line to the curve at (8, 2) is given by the derivative \frac{dy}{dx} at that point. By the chain rule,

\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}

Differentiate the given parametric equations with respect to t :

x = 4t \implies \dfrac{dx}{dt} = 4

y = \dfrac4t \implies \dfrac{dy}{dt} = -\dfrac4{t^2}

Then

\dfrac{dy}{dx} = \dfrac{-\frac4{t^2}}4 = -\dfrac1{t^2}

We have x=8 and y=2 when t=2, so the slope at the given point is \frac{dy}{dx} = -\frac14.

The normal line to the same point is perpendicular to the tangent line, so its slope is +4. Then using the point-slope formula for a line, the normal line has equation

y - 2 = 4 (x - 8) \implies \boxed{y = 4x - 30}

Alternatively, we can eliminate the parameter and express y explicitly in terms of x :

x = 4t \implies t = \dfrac x4 \implies y = \dfrac4t = \dfrac4{\frac x4} = \dfrac{16}x

Then the slope of the tangent line is

\dfrac{dy}{dx} = -\dfrac{16}{x^2}

At x = 8, the slope is again -\frac{16}{64}=-\frac14, so the normal has slope +4, and so on.

5 0
2 years ago
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