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3 years ago

8

What is the area of a triangle with a base of 7 cm and a height of 4cm

1 answer:

Alex777 [14]3 years ago

3 0

Answer:

14 sq cm

Step-by-step explanation:

7 × 4 = 28

28 ÷ 2 = 14

brainliest?

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The condition_______?proves that ∆ABC and ∆EFG are congruent by the SAS criterion.

snow_lady [41]

Answer:

(1) D.Angle C is congruent to to Angle F. (2) C. SSS. (3) C. cannot be congruent to.

Step-by-step explanation:

1)

From the given figure it is noticed that

$AC=EG$

$CB=GF$

According to SAS postulate, if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then both triangles are congruent.

The included angles of congruent sides are angle C and angle G.

So, condition "Angle C is congruent to to Angle F" will prove that the ∆ABC and ∆EFG are congruent by the SAS criterion.

2)

If $AB\neq EF$

According to SSS postulate, if all three sides in one triangle are congruent to the corresponding sides in the other.

Since two corresponding sides are congruent but third sides of triangles are not congruent, therefore SSS criterion for congruence is violated.

3)

Since two corresponding sides are congruent but third sides of triangles are not congruent, therefore the included angle of congruent sides are different.

$\angle C\neq \angle G$

Therefore angle C and angle F cannot be congruent to each other.

4 0

3 years ago

Use Hooke's Law to determine the work done by the variable force in the spring problem. A force of 450 newtons stretches a sprin

vlada-n [284]

Answer:

The work done is 202.50Nm

Step-by-step explanation:

Given

$F =450N$

$x_1 = 30cm$

$x_2 = 60cm$

Required

The work done

First, we calculate the spring constant (k)

$F = kx_1$

$450N = k *30cm$

$k = \frac{450N}{30cm}$

$k =15N/cm$

So:

$F = kx_1$

$F(x) = 15x$

The work done using Hooke's law is:

$W =\int\limits^a_b {F(x)} \, dx$

This gives:

$W =\int\limits^{60}_{30} {15x} \, dx$

Rewrite as:

$W =15\int\limits^{60}_{30} {x} \, dx$

Integrate

$W =15 \frac{x^2}{2}|\limits^{60}_{30}$

This gives:

$W =15 *\frac{60^2 - 30^2}{2}$

$W =15 *\frac{2700}{2}$

$W =15 *1350$

$W =20250N-cm$

Convert to Nm

$W =\frac{20250Nm}{100}$

$W =202.50Nm$

7 0

3 years ago

The width of the rectangle is 75% of its length, the length is 24, what's the area?

lapo4ka [179]

S = a * b where a - <span>length and b - width
a = 24
b = 0.75 * a
S = 24 * 24 * 0.75 = 432</span>

7 0

3 years ago

Read 2 more answers

301.6=3.14•4•h/3 find h

skad [1K]

301.6=3.14×4×h/3
301.6=12.56×h/3
3770/157=h/3
72 6/157=h. Hope it help!

3 0

4 years ago

Adult male heights have a normal probability distribution with a mean of 70 inches and a standard deviation of 4 inches. What is

raketka [301]

Answer: 0.5

Step-by-step explanation:

Given : Adult male heights have a normal probability distribution .

Population mean : $\mu = 70 \text{ inches}$

Standard deviation: $\sigma= 4\text{ inches}$

Let x be the random variable that represent the heights of adult male.

z-score : $z=\dfrac{x-\mu}{\sigma}$

For x=70, we have

$z=\dfrac{70-70}{4}=0$

Now, by using the standard normal distribution table, we have

The probability that a randomly selected male is more than 70 inches tall :-

$P(x\geq70)=P(z\geq0)=1-P(z$

Hence, the probability that a randomly selected male is more than 70 inches tall = 0.5

4 0

3 years ago

Read 2 more answers