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Thepotemich [5.8K]
2 years ago
8

What value is equivalent to -30-2^3*3​

Mathematics
2 answers:
zepelin [54]2 years ago
8 0

Answer:

-6

Step-by-step explanation:

-2 ^3x3 = 24

-30 + 24 = -6

AfilCa [17]2 years ago
5 0

- 30 - 2³ × 3

= - 30 - 8 × 3

= - 30 - 24

= - 54

Follow the order of PEMDAS.

P =》Parenthesis

E =》Exponents

M =》Multiplication

D =》Division

A =》Addition

S =》Subtraction

____

Hope it helps ⚜

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2 years ago
T what point does the curve have maximum curvature? Y = 7ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as
Nookie1986 [14]

Formula for curvature for a well behaved curve y=f(x) is


K(x)= \frac{|{y}''|}{[1+{y}'^2]^\frac{3}{2}}


The given curve is y=7e^{x}


{y}''=7e^{x}\\ {y}'=7e^{x}


k(x)=\frac{7e^{x}}{[{1+(7e^{x})^2}]^\frac{3}{2}}


{k(x)}'=\frac{7(e^x)(1+49e^{2x})(49e^{2x}-\frac{1}{2})}{[1+49e^{2x}]^{3}}

For Maxima or Minima

{k(x)}'=0

7(e^x)(1+49e^{2x})(98e^{2x}-1)=0

→e^{x}=0∨ 1+49e^{2x}=0∨98e^{2x}-1=0

e^{x}=0  ,  ∧ 1+49e^{2x}=0   [not possible ∵there exists no value of x satisfying these equation]

→98e^{2x}-1=0

Solving this we get

x= -\frac{1}{2}\ln{98}

As you will evaluate {k(x})}''<0 at x=-\frac{1}{2}\ln98

So this is the point of Maxima. we get y=7×1/√98=1/√2

(x,y)=[-\frac{1}{2}\ln98,1/√2]

k(x)=\lim_{x\to\infty } \frac{7e^{x}}{[{1+(7e^{x})^2}]^\frac{3}{2}}

k(x)=\frac{7}{\infty}

k(x)=0







5 0
3 years ago
PLEASE HELP SOLVE FOR ‘X’!!!
antoniya [11.8K]

Answer:

x= 10

Step-by-step explanation:

1/5 x -2/3 = 4/3

Add 2/3 to each side

1/5x -2/3 +2/3 = 4/3 +2/3

1/5x = 6/3

1/5x = 2

Multiply each side by 5

1/5x * 5 = 2*5

x = 10

3 0
2 years ago
Express the product in simplest form.
Dmitry [639]

The product in simplest form is (x - 4)

<em><u>Solution:</u></em>

<em><u>Given expression is:</u></em>

\frac{8}{2x+8} \times \frac{x^2-16}{4}

We have to find the product in simplest form

In the given expression,

2x + 8 = 2(x+ 4)

We know that,

a^2-b^2=(a+b)(a-b)

Therefore,

x^2-16 = x^2-4^2 =(x+4)(x-4)

Substitute these in given expression

\frac{8}{2(x+4)} \times \frac{(x+4)(x-4)}{4}

Cancel the common factors,

\frac{8}{2(x+4)} \times \frac{(x+4)(x-4)}{4} = x - 4

Thus the product in simplest form is (x - 4)

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Write a polynomial function with rational coefficients so that P(x)=0 has the given root.
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(x-7)*(x-12)= x^2-19x+84
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