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3 years ago

Question

Mathematics

1 answer:

nata0808 [166]3 years ago

5 0

Answer:5x

Step-by-step explanation:

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The area of the base of a rectangular prisim four and three fourths and the height of two and one third what is the volume

VikaD [51]

Data:
$A_{base} = 4 \frac{3}{4} = \frac{4*4+3}{4} = \frac{16+3}{4} \to A_{base} = \frac{19}{4}$

$h (height) = 2 \frac{1}{3} = \frac{3*2+1}{3} = \frac{6+1}{3} \to h (height) = \frac{7}{3}$
$v (volume) = ?$

Formula: $V = A_{b}*h$

Solving:
$V = A_{b}*h$
$V = \frac{19}{4} * \frac{7}{3}$
$V = \frac{133}{12}$
$V = 11.08333....$
$\boxed{\boxed{V \approx 11.08}}\end{array}}\qquad\quad\checkmark$

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3 years ago

Which combination of integers can be used to generate the Pythagorean triple (7, 24, 25)

amm1812

Answer:

Step-by-step explanation:

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2 years ago

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Question- Solve for a 2a+b=x

kobusy [5.1K]

Answer:

a = (x - b)/2

Step-by-step explanation:

2a + b = x

-b -b

2a = x - b

/2 /2

a = (x - b)/2

for ex:

x = 3

b = 2

a = ((3) - (2))/2 = 0.5

a = 0.5

2a + b = x

2(0.5) + 2 = 3

4 0

2 years ago

Please answer : ) will give brainly go to last problem for it to make sense

Ede4ka [16]

Answer:

30%

Step-by-step explanation:

if 60% were thin mint and 10% were Samoas that is 70% of cookies. remaining 30% must be trefoils. hope this helps

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2 years ago

1+secA/sec A = sin^2 A / 1-cos A

Fofino [41]

Answer: see proof below

Step-by-step explanation:

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

Use the following Identities:

sec Ф = 1/cos Ф

cos² Ф + sin² Ф = 1

Proof LHS → RHS

$\text{LHS:}\qquad \qquad \dfrac{1+\sec A}{\sec A}$

$\text{Identity:}\qquad \qquad \dfrac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}$

$\text{Simplify:}\qquad \qquad \dfrac{\frac{\cos A+1}{\cos A}}{\frac{1}{\cos A}}\\\\\\.\qquad \qquad \qquad =\dfrac{1+\cos A}{1}$

$\text{Multiply:}\qquad \qquad \dfrac{1+\cos A}{1}\cdot \bigg(\dfrac{1-\cos A}{1-\cos A}\bigg)\\\\\\.\qquad \qquad \qquad =\dfrac{1-\cos^2 A}{1-\cos A}$

$\text{Identity:}\qquad \qquad \dfrac{\sin^2 A}{1-\cos A}$

$\text{LHS = RHS:}\quad \dfrac{\sin^2 A}{1-\cos A}=\dfrac{\sin^2 A}{1-\cos A}\quad \checkmark$

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3 years ago