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2 years ago

What is the answer to this please help!

Mathematics

1 answer:

Butoxors [25]2 years ago

6 0

$z+(-0.72)=-2.5\\\\\text{Step 1}:\ x+(-0.72)+(-0.72)=-2.5+(-0.72)\qquad\text{ERROR}\\\\CORRECT\ IS:\ x+(-0.72)-(-0.72)=-2.5-(-0.72)\\\\\text{Step 2}:\ x=-2.5+0.72\\\\\text{Step 3}:\ x=-1.78$

Subtract, not add, -0.72 from both sides.

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Please help with this square problem, the picture is shown.

Galina-37 [17]

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Total area = 10x + 6x + 36 = 16x + 36 yd²

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2 years ago

A football team gained 10 yards on one play and then lost 22 yards on the next. What was the overall change in field position.

grandymaker [24]

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2 years ago

Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1

sleet_krkn [62]

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}$

From Left side:

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)$

$\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}$

NOTE: sin²θ + cos²θ = 1

$\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}$

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$\dfrac{2sec^2\theta}{tan^2\theta-1}$

Left side = Right side so proof is complete

8 0

2 years ago

Read 2 more answers

Last week, Clay wrote 4 checks for $20 each from his checking account., withdrew another $30, and finally made a deposit of $96.

Aliun [14]

(4x20)+30=$110 taken out. if he had $96 put back in, then go $110-$96, leaving a balance of -$14, or in other words, the balance dropped $14.

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2 years ago

Give the domain and range of the relation.

Alex_Xolod [135]

Answer:

Please check the explanation.

Step-by-step explanation:

Finding Domain:

We know that the domain of a function is the set of input or argument values for which the function is real and defined.

From the given graph, it is clear that the starting x-value of the line is x=-2, the closed circle at the starting value of x= -2 means the x-value x=-2 is included.

And the line ends at the x-value x=1 with a closed circle, meaning the ending value of x=1 is also included.

Thus, the domain is:

D: {-2, -1, 0, 1} or D: −2 ≤ x ≤ 1

Finding Range:

We also know that the range of a function is the set of values of the dependent variable for which a function is defined

From the given graph, it is clear that the starting y-value of the line is y=0, the closed circle at the starting value of y = 0 means the y-value y=0 is included.

And the line ends at the y-value y=2 with a closed circle, meaning the ending value of y=2 is also included.

Thus, the range is:

R: {0, 1, 2} or R: 0 ≤ y ≤ 2

7 0

2 years ago