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3 years ago

PLZ HELP I HAVE NO CLUE WHAT TO DO

Mathematics

1 answer:

EleoNora [17]3 years ago

5 0

Answer:

142 cm^2

Step-by-step explanation:

2(L*H)+2(L*W)+2(W*H)=SA

7*3=21

21*2=42

7*5=35

35*2=70

5*3=15*2=30

30+70+42=142

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Figure abcd is parallelogram. What are measures of B and D?

nika2105 [10]

Answer:

Both are 55

Step-by-step explanation:

To solve this problem, set the equations equal to each other, solve for n, then plug in the values

Solve for n:

1. 2n+15=3n-5

2. Add 5 to both sides

3. Now you have 2n+20=3n

4. Subtract 2n from both sides

5. You are left with n=20

Next, plug it in:

1. Replace all n's with 20

2. 2(20)+15=55

3. 3(20)-5=55

There you go! Hope this helps!

3 0

3 years ago

Please help ! I have been stuck on this calculus problems, I will mark you brainliest!

Anna71 [15]

Answer:

$\displaystyle J'(3) = -1$

General Formulas and Concepts:

Algebra I

Functions
Function Notation

Calculus

Derivatives

Derivative Notation

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Derivative: $\displaystyle \frac{d}{dx} [e^u]=e^u \cdot u'$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle J(x) = e^{f(x)}$

Step 2: Differentiate

eˣ Derivative [Derivative Rule - Chain Rule]: $\displaystyle J'(x) = \frac{d}{dx}[e^{f(x)}] \cdot \frac{d}{dx}[f(x)]$
Simplify: $\displaystyle J'(x) = f'(x)e^{f(x)}$

Step 3: Evaluate

Substitute in x [Derivative]: $\displaystyle J'(3) = f'(3)e^{f(3)}$
Substitute in function values: $\displaystyle J'(3) = -e^{0}$
Simplify: $\displaystyle J'(3) = -1$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Derivatives

Book: College Calculus 10e

4 0

3 years ago

Is the function negative over (-6,-2)

MakcuM [25]

Yes, the function would be negative

5 0

3 years ago

Solve the simultaneous equation y-2x+1=0 and 4x^2+3y^2-2xy=7

Lapatulllka [165]

The first equation is $y-2x+1=0$

$y=2x-1$ (Equation 1)

The second equation is $4x^{2}+3y^{2}-2xy=7$ (Equation 2)

Putting the value of x from equation 1 in equation 2.

we get,

$4x^{2}+3(2x-1)^{2}-2x(2x-1)=7$

$4x^{2}+3(4x^{2}+1-4x)-4x^{2}+2x=7$

by simplifying the given equation,

$12x^{2}-10x-4=0$

$6x^{2}-5x-2=0$

Using discriminant formula,

$D=b^{2}-4ac$

$D=25-4 \times 6 \times -2 = 73$

Now the formula for solution 'x' of quadratic equation is given by:

$x=\frac{-b+\sqrt{D}}{2a} and x=\frac{-b-\sqrt{D}}{2a}$

$x=\frac{5+\sqrt{73}}{12} and x=\frac{5-\sqrt{73}}{12}$

Hence, these are the required solutions.

8 0

3 years ago

Find the first four terms of the multiplication pattern given by the formula.

lara31 [8.8K]

You just need to plug in 1, 2, 3, and 4.
10*4(1)-1=39
10*4(2)-1=79
10*4(3)-1=119
10*4(4)-1=159

6 0

3 years ago