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3 years ago

Help me please,Thanks

Mathematics

1 answer:

Natali5045456 [20]3 years ago

5 0

Area of triangle is 1/2 b times h. So do 12 times 10= 120 then do 120/2 which equals 60

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A line passes through (2, −1) and (4, 5).

viktelen [127]

the answer to your question is
−3x+y=−7

5 0

3 years ago

What are the coefficients in the expression x-3y-7xy+4?

IRINA_888 [86]

The coefficients in the expression x-3y-7xy+4 is -3 and -7.

The coefficient of x are 1

-3y=3

-7xy=-7

What is coefficients?

A coefficient is a number or quantity that is associated with a variable. Typically, it is an integer multiplied by the variable and written next to it. The variables which do not have a number with them are assumed to be having 1 as their coefficient. For example, 3 is the coefficient of x in the expression 3x, but 1 is the coefficient of x2 in the expression x2 + 3. A coefficient, in other words, is a multiplicative factor in terms of a polynomial, a series, or any expression. Consider the following expression, which shows that 5 is the x2 coefficient and 8 is the y coefficient.

To learn more about coefficients from the given link:

brainly.com/question/27481600

#SPJ9

8 0

2 years ago

Can you please help me solve this??

fredd [130]

Answer:

Step-by-step explanation:

1.25 yd x 2 yd x 1.5 yd

= 2.5 yd^2 x 1.5 yd

= 3 3/4 yd^3

answer is D

7 0

3 years ago

Evaluated to square

denis-greek [22]

Step-by-step explanation:

The square root of a number is a number that you can square to get it, that is, a number that you can multiply by itself to get the number. So, 2 is a square root of 4, because 2 x 2 = 4, and 3 is a square root of 9, because 3 x 2 = 9.

3 0

3 years ago

Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = 2 cos(t) + sin(2t),[0, π/2].

castortr0y [4]

Answer:

The absolute maximum is $\frac{3\sqrt 3}2$ and the absolute minimum value is $0.$

Step-by-step explanation:

Differentiate of $f$ both sides w.r.t. $t$ ,

$f(t)=2 \cos t+\sin 2t$

$\Rightarrow f'(t)=-2\sin t+2\cos 2t$

Now take $f'(t)=0$

$\Rightarrow -2\sin t+2\cos 2t=0$

$\Rightarrow 2\cos 2t=2\sin t$

$\Rightarrow \cos 2t=\sin t$

$\Rightarrow 1-2\sin ^2t =\sin t \quad \quad [\because \cos 2t = 1-2\sin ^2t]$

$\Rightarrow 2\sin ^2t+\sin t-1=0$

$\Rightarrow 2\sin ^2t+2\sin t-\sin t-1=0$

$\Rightarrow 2\sin t(\sin t+1)-1(\sin t+1)=0$

$\Rightarrow (\sin t+1)(2\sin t-1)=0$

$\Rightarrow \sin t+1=0 \;\text{and}\; 2\sin t-1=0$

$\Rightarrow \sin t =-1 \;\text{and}\; \sin t =\frac 12$

In the interval $0\leq t\leq \frac {\pi}2$ , the answer to this problem is $\frac {\pi}6$

Now find the second derivative of $f(t)$ w.r.t. $t$ ,

$f''(t)=-2\cos t-4\sin 2t$

$\Rightarrow \left[f''(t)\right]_{t=\frac {\pi}6}=-2\times \frac {\sqrt 3}2-4\times \frac{\sqrt 3}2=-3\sqrt 3$

Thus, $f(t)$ is maximum at $t=\frac {\pi}6$ and minimum at $t=0$

$\left[f(t)\right]_{t=\frac {\pi}6}=2\times \frac {\sqrt 3}2+\frac{\sqrt 3}2=\frac{3\sqrt 3}2\;\text{and}\;\left[f(t)\right]_{t=\frac{\pi}2}= 2\times 0+0=0$

Hence, the absolute maximum is $\frac{3\sqrt 3}2$ and the absolute minimum value is $0$ .

7 0

3 years ago